work in progress; to be tidied and fixed

[u/mdw/putty] / sshbn.c

/*
 * Bignum routines for RSA and DH and stuff.
 */

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>

#include "misc.h"
#include "bn-internal.h"
#include "ssh.h"

BignumInt bnZero[1] = { 0 };
BignumInt bnOne[2] = { 1, 1 };

/*
 * The Bignum format is an array of `BignumInt'. The first
 * element of the array counts the remaining elements. The
 * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
 * significant digit first. (So it's trivial to extract the bit
 * with value 2^n for any n.)
 *
 * All Bignums in this module are positive. Negative numbers must
 * be dealt with outside it.
 *
 * INVARIANT: the most significant word of any Bignum must be
 * nonzero.
 */

Bignum Zero = bnZero, One = bnOne;

static Bignum newbn(int length)
{
    Bignum b = snewn(length + 1, BignumInt);
    if (!b)
        abort();                       /* FIXME */
    memset(b, 0, (length + 1) * sizeof(*b));
    b[0] = length;
    return b;
}

void bn_restore_invariant(Bignum b)
{
    while (b[0] > 1 && b[b[0]] == 0)
        b[0]--;
}

Bignum copybn(Bignum orig)
{
    Bignum b = snewn(orig[0] + 1, BignumInt);
    if (!b)
        abort();                       /* FIXME */
    memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
    return b;
}

void freebn(Bignum b)
{
    /*
     * Burn the evidence, just in case.
     */
    smemclr(b, sizeof(b[0]) * (b[0] + 1));
    sfree(b);
}

Bignum bn_power_2(int n)
{
    Bignum ret = newbn(n / BIGNUM_INT_BITS + 1);
    bignum_set_bit(ret, n, 1);
    return ret;
}

/*
 * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
 * little-endian arrays of 'len' BignumInts. Returns a BignumInt carried
 * off the top.
 */
static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
                              BignumInt *c, int len)
{
    int i;
    BignumDblInt carry = 0;

    for (i = 0; i < len; i++) {
        carry += (BignumDblInt)a[i] + b[i];
        c[i] = (BignumInt)carry;
        carry >>= BIGNUM_INT_BITS;
    }

    return (BignumInt)carry;
}

/*
 * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
 * all little-endian arrays of 'len' BignumInts. Any borrow from the top
 * is ignored.
 */
static void internal_sub(const BignumInt *a, const BignumInt *b,
                         BignumInt *c, int len)
{
    int i;
    BignumDblInt carry = 1;

    for (i = 0; i < len; i++) {
        carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
        c[i] = (BignumInt)carry;
        carry >>= BIGNUM_INT_BITS;
    }
}

/*
 * Compute c = a * b.
 * Input is in the first len words of a and b.
 * Result is returned in the first 2*len words of c.
 *
 * 'scratch' must point to an array of BignumInt of size at least
 * mul_compute_scratch(len). (This covers the needs of internal_mul
 * and all its recursive calls to itself.)
 */
#define KARATSUBA_THRESHOLD 50
static int mul_compute_scratch(int len)
{
    int ret = 0;
    while (len > KARATSUBA_THRESHOLD) {
        int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
        int midlen = botlen + 1;
        ret += 4*midlen;
        len = midlen;
    }
    return ret;
}
static void internal_mul(const BignumInt *a, const BignumInt *b,
                         BignumInt *c, int len, BignumInt *scratch)
{
    if (len > KARATSUBA_THRESHOLD) {
        int i;

        /*
         * Karatsuba divide-and-conquer algorithm. Cut each input in
         * half, so that it's expressed as two big 'digits' in a giant
         * base D:
         *
         *   a = a_1 D + a_0
         *   b = b_1 D + b_0
         *
         * Then the product is of course
         *
         *  ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
         *
         * and we compute the three coefficients by recursively
         * calling ourself to do half-length multiplications.
         *
         * The clever bit that makes this worth doing is that we only
         * need _one_ half-length multiplication for the central
         * coefficient rather than the two that it obviouly looks
         * like, because we can use a single multiplication to compute
         *
         *   (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
         *
         * and then we subtract the other two coefficients (a_1 b_1
         * and a_0 b_0) which we were computing anyway.
         *
         * Hence we get to multiply two numbers of length N in about
         * three times as much work as it takes to multiply numbers of
         * length N/2, which is obviously better than the four times
         * as much work it would take if we just did a long
         * conventional multiply.
         */

        int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
        int midlen = botlen + 1;
        BignumDblInt carry;

        /*
         * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
         * in the output array, so we can compute them immediately in
         * place.
         */

#ifdef KARA_DEBUG
        printf("a1,a0 = 0x");
        for (i = 0; i < len; i++) {
            if (i == toplen) printf(", 0x");
            printf("%0*x", BIGNUM_INT_BITS/4, a[len - 1 - i]);
        }
        printf("\n");
        printf("b1,b0 = 0x");
        for (i = 0; i < len; i++) {
            if (i == toplen) printf(", 0x");
            printf("%0*x", BIGNUM_INT_BITS/4, b[len - 1 - i]);
        }
        printf("\n");
#endif

        /* a_1 b_1 */
        internal_mul(a + botlen, b + botlen, c + 2*botlen, toplen, scratch);
#ifdef KARA_DEBUG
        printf("a1b1 = 0x");
        for (i = 0; i < 2*toplen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, c[2*len - 1 - i]);
        }
        printf("\n");
#endif

        /* a_0 b_0 */
        internal_mul(a, b, c, botlen, scratch);
#ifdef KARA_DEBUG
        printf("a0b0 = 0x");
        for (i = 0; i < 2*botlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, c[2*botlen - 1 - i]);
        }
        printf("\n");
#endif

        /* Zero padding. botlen exceeds toplen by at most 1, and we'll set
         * the extra carry explicitly below, so we only need to zero at most
         * one of the top words here.
         */
        scratch[midlen - 2] = scratch[2*midlen - 2] = 0;

        for (i = 0; i < toplen; i++) {
            scratch[i] = a[i + botlen]; /* a_1 */
            scratch[midlen + i] = b[i + botlen]; /* b_1 */
        }

        /* compute a_1 + a_0 */
        scratch[midlen - 1] = internal_add(scratch, a, scratch, botlen);
#ifdef KARA_DEBUG
        printf("a1plusa0 = 0x");
        for (i = 0; i < midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen - 1 - i]);
        }
        printf("\n");
#endif
        /* compute b_1 + b_0 */
        scratch[2*midlen - 1] = internal_add(scratch+midlen, b,
                                             scratch+midlen, botlen);
#ifdef KARA_DEBUG
        printf("b1plusb0 = 0x");
        for (i = 0; i < midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen - 1 - i]);
        }
        printf("\n");
#endif

        /*
         * Now we can do the third multiplication.
         */
        internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
                     scratch + 4*midlen);
#ifdef KARA_DEBUG
        printf("a1plusa0timesb1plusb0 = 0x");
        for (i = 0; i < 2*midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[4*midlen - 1 - i]);
        }
        printf("\n");
#endif

        /*
         * Now we can reuse the first half of 'scratch' to compute the
         * sum of the outer two coefficients, to subtract from that
         * product to obtain the middle one.
         */
        scratch[2*botlen - 2] = scratch[2*botlen - 1] = 0;
        for (i = 0; i < 2*toplen; i++)
            scratch[i] = c[2*botlen + i];
        scratch[2*botlen] = internal_add(scratch, c, scratch, 2*botlen);
        scratch[2*botlen + 1] = 0;
#ifdef KARA_DEBUG
        printf("a1b1plusa0b0 = 0x");
        for (i = 0; i < 2*midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen - 1 - i]);
        }
        printf("\n");
#endif

        internal_sub(scratch + 2*midlen, scratch, scratch, 2*midlen);
#ifdef KARA_DEBUG
        printf("a1b0plusa0b1 = 0x");
        for (i = 0; i < 2*midlen; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, scratch[4*midlen - 1 - i]);
        }
        printf("\n");
#endif

        /*
         * And now all we need to do is to add that middle coefficient
         * back into the output. We may have to propagate a carry
         * further up the output, but we can be sure it won't
         * propagate right the way off the top.
         */
        carry = internal_add(c + botlen, scratch, c + botlen, 2*midlen);
        i = botlen + 2*midlen;
        while (carry) {
            assert(i <= 2*len);
            carry += c[i];
            c[i] = (BignumInt)carry;
            carry >>= BIGNUM_INT_BITS;
            i++;
        }
#ifdef KARA_DEBUG
        printf("ab = 0x");
        for (i = 0; i < 2*len; i++) {
            printf("%0*x", BIGNUM_INT_BITS/4, c[2*len - i]);
        }
        printf("\n");
#endif

    } else {
        int i;
        BignumInt carry;
        BignumDblInt t;
        const BignumInt *ap, *alim = a + len, *bp, *blim = b + len;
        BignumInt *cp, *cps;

        /*
         * Multiply in the ordinary O(N^2) way.
         */

        for (i = 0; i < 2 * len; i++)
            c[i] = 0;

        for (cps = c, ap = a; ap < alim; ap++, cps++) {
            carry = 0;
            for (cp = cps, bp = b, i = blim - bp; i--; bp++, cp++) {
                t = (MUL_WORD(*ap, *bp) + carry) + *cp;
                *cp = (BignumInt) t;
                carry = (BignumInt)(t >> BIGNUM_INT_BITS);
            }
            *cp = carry;
        }
    }
}

/*
 * Variant form of internal_mul used for the initial step of
 * Montgomery reduction. Only bothers outputting 'len' words
 * (everything above that is thrown away).
 */
static void internal_mul_low(const BignumInt *a, const BignumInt *b,
                             BignumInt *c, int len, BignumInt *scratch)
{
    if (len > KARATSUBA_THRESHOLD) {
        int i;

        /*
         * Karatsuba-aware version of internal_mul_low. As before, we
         * express each input value as a shifted combination of two
         * halves:
         *
         *   a = a_1 D + a_0
         *   b = b_1 D + b_0
         *
         * Then the full product is, as before,
         *
         *  ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
         *
         * Provided we choose D on the large side (so that a_0 and b_0
         * are _at least_ as long as a_1 and b_1), we don't need the
         * topmost term at all, and we only need half of the middle
         * term. So there's no point in doing the proper Karatsuba
         * optimisation which computes the middle term using the top
         * one, because we'd take as long computing the top one as
         * just computing the middle one directly.
         *
         * So instead, we do a much more obvious thing: we call the
         * fully optimised internal_mul to compute a_0 b_0, and we
         * recursively call ourself to compute the _bottom halves_ of
         * a_1 b_0 and a_0 b_1, each of which we add into the result
         * in the obvious way.
         *
         * In other words, there's no actual Karatsuba _optimisation_
         * in this function; the only benefit in doing it this way is
         * that we call internal_mul proper for a large part of the
         * work, and _that_ can optimise its operation.
         */

        int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */

        /*
         * Scratch space for the various bits and pieces we're going
         * to be adding together: we need botlen*2 words for a_0 b_0
         * (though we may end up throwing away its topmost word), and
         * toplen words for each of a_1 b_0 and a_0 b_1. That adds up
         * to exactly 2*len.
         */

        /* a_0 b_0 */
        internal_mul(a, b, scratch + 2*toplen, botlen, scratch + 2*len);

        /* a_1 b_0 */
        internal_mul_low(a + botlen, b, scratch + toplen, toplen,
                         scratch + 2*len);

        /* a_0 b_1 */
        internal_mul_low(a, b + botlen, scratch, toplen, scratch + 2*len);

        /* Copy the bottom half of the big coefficient into place */
        for (i = 0; i < botlen; i++)
            c[i] = scratch[2*toplen + i];

        /* Add the two small coefficients, throwing away the returned carry */
        internal_add(scratch, scratch + toplen, scratch, toplen);

        /* And add that to the large coefficient, leaving the result in c. */
        internal_add(scratch, scratch + 2*toplen + botlen,
                     c + botlen, toplen);

    } else {
        int i;
        BignumInt carry;
        BignumDblInt t;
        const BignumInt *ap, *alim = a + len, *bp;
        BignumInt *cp, *cps, *clim = c + len;

        /*
         * Multiply in the ordinary O(N^2) way.
         */

        for (i = 0; i < len; i++)
            c[i] = 0;

        for (cps = c, ap = a; ap < alim; ap++, cps++) {
            carry = 0;
            for (cp = cps, bp = b, i = clim - cp; i--; bp++, cp++) {
                t = (MUL_WORD(*ap, *bp) + carry) + *cp;
                *cp = (BignumInt) t;
                carry = (BignumInt)(t >> BIGNUM_INT_BITS);
            }
        }
    }
}

/*
 * Montgomery reduction. Expects x to be a little-endian array of 2*len
 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
 * x' < n.
 *
 * 'n' and 'mninv' should be little-endian arrays of 'len' BignumInts
 * each, containing respectively n and the multiplicative inverse of
 * -n mod r.
 *
 * 'tmp' is an array of BignumInt used as scratch space, of length at
 * least 3*len + mul_compute_scratch(len).
 */
static void monty_reduce(BignumInt *x, const BignumInt *n,
                         const BignumInt *mninv, BignumInt *tmp, int len)
{
    int i;
    BignumInt carry;

    /*
     * Multiply x by (-n)^{-1} mod r. This gives us a value m such
     * that mn is congruent to -x mod r. Hence, mn+x is an exact
     * multiple of r, and is also (obviously) congruent to x mod n.
     */
    internal_mul_low(x, mninv, tmp, len, tmp + 3*len);

    /*
     * Compute t = (mn+x)/r in ordinary, non-modular, integer
     * arithmetic. By construction this is exact, and is congruent mod
     * n to x * r^{-1}, i.e. the answer we want.
     *
     * The following multiply leaves that answer in the _most_
     * significant half of the 'x' array, so then we must shift it
     * down.
     */
    internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
    carry = internal_add(x, tmp+len, x, 2*len);
    for (i = 0; i < len; i++)
        x[i] = x[len + i], x[len + i] = 0;

    /*
     * Reduce t mod n. This doesn't require a full-on division by n,
     * but merely a test and single optional subtraction, since we can
     * show that 0 <= t < 2n.
     *
     * Proof:
     *  + we computed m mod r, so 0 <= m < r.
     *  + so 0 <= mn < rn, obviously
     *  + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
     *  + yielding 0 <= (mn+x)/r < 2n as required.
     */
    if (!carry) {
        for (i = len; i-- > 0; )
            if (x[i] != n[i])
                break;
    }
    if (carry || i < 0 || x[i] > n[i])
        internal_sub(x, n, x, len);
}

static void internal_add_shifted(BignumInt *number,
                                 unsigned n, int shift)
{
    int word = 1 + (shift / BIGNUM_INT_BITS);
    int bshift = shift % BIGNUM_INT_BITS;
    BignumDblInt addend;

    addend = (BignumDblInt)n << bshift;

    while (addend) {
        addend += number[word];
        number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
        addend >>= BIGNUM_INT_BITS;
        word++;
    }
}

/*
 * Compute a = a % m.
 * Input in first alen words of a and first mlen words of m.
 * Output in first alen words of a
 * (of which last alen-mlen words will be zero).
 * The MSW of m MUST have its high bit set.
 * Quotient is accumulated in the `quotient' array. Quotient parts
 * are shifted left by `qshift' before adding into quot.
 */
static void internal_mod(BignumInt *a, int alen,
                         BignumInt *m, int mlen,
                         BignumInt *quot, int qshift)
{
    BignumInt m0, m1;
    unsigned int h;
    int i, j, k;

    m0 = m[mlen - 1];
    if (mlen > 1)
        m1 = m[mlen - 2];
    else
        m1 = 0;

    for (i = alen, h = 0; i-- >= mlen; ) {
        BignumDblInt t;
        unsigned int q, r, c, ai1;

        if (i)
            ai1 = a[i - 1];
        else
            ai1 = 0;

        /* Find q = h:a[i] / m0 */
        if (h >= m0) {
            /*
             * Special case.
             * 
             * To illustrate it, suppose a BignumInt is 8 bits, and
             * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
             * our initial division will be 0xA123 / 0xA1, which
             * will give a quotient of 0x100 and a divide overflow.
             * However, the invariants in this division algorithm
             * are not violated, since the full number A1:23:... is
             * _less_ than the quotient prefix A1:B2:... and so the
             * following correction loop would have sorted it out.
             * 
             * In this situation we set q to be the largest
             * quotient we _can_ stomach (0xFF, of course).
             */
            q = BIGNUM_INT_MASK;
        } else {
            /* Macro doesn't want an array subscript expression passed
             * into it (see definition), so use a temporary. */
            BignumInt tmplo = a[i];
            DIVMOD_WORD(q, r, h, tmplo, m0);

            /* Refine our estimate of q by looking at
             h:a[i]:a[i-1] / m0:m1 */
            t = MUL_WORD(m1, q);
            if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
                q--;
                t -= m1;
                r = (r + m0) & BIGNUM_INT_MASK;     /* overflow? */
                if (r >= (BignumDblInt) m0 &&
                    t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
            }
        }

        j = i + 1 - mlen;

        /* Subtract q * m from a[i...] */
        c = 0;
        for (k = 0; k < mlen; k++) {
            t = MUL_WORD(q, m[k]);
            t += c;
            c = (unsigned)(t >> BIGNUM_INT_BITS);
            if ((BignumInt) t > a[j + k])
                c++;
            a[j + k] -= (BignumInt) t;
        }

        /* Add back m in case of borrow */
        if (c != h) {
            t = 0;
            for (k = 0; k < mlen; k++) {
                t += m[k];
                t += a[j + k];
                a[j + k] = (BignumInt) t;
                t = t >> BIGNUM_INT_BITS;
            }
            q--;
        }

        if (quot)
            internal_add_shifted(quot, q,
                                 qshift + BIGNUM_INT_BITS * (i + 1 - mlen));

        if (i >= mlen) {
            h = a[i];
            a[i] = 0;
        }
    }
}

static void shift_left(BignumInt *x, int xlen, int shift)
{
    int i;

    if (!shift)
        return;
    for (i = xlen; --i > 0; )
        x[i] = (x[i] << shift) | (x[i - 1] >> (BIGNUM_INT_BITS - shift));
    x[0] = x[0] << shift;
}

static void shift_right(BignumInt *x, int xlen, int shift)
{
    int i;

    if (!shift || !xlen)
        return;
    xlen--;
    for (i = 0; i < xlen; i++)
        x[i] = (x[i] >> shift) | (x[i + 1] << (BIGNUM_INT_BITS - shift));
    x[i] = x[i] >> shift;
}

/*
 * Compute (base ^ exp) % mod, the pedestrian way.
 */
Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
    BignumInt *a, *b, *n, *m, *scratch;
    int mshift;
    int mlen, scratchlen, i, j;
    Bignum base, result;

    /*
     * The most significant word of mod needs to be non-zero. It
     * should already be, but let's make sure.
     */
    assert(mod[mod[0]] != 0);

    /*
     * Make sure the base is smaller than the modulus, by reducing
     * it modulo the modulus if not.
     */
    base = bigmod(base_in, mod);

    /* Allocate m of size mlen, copy mod to m */
    mlen = mod[0];
    m = snewn(mlen, BignumInt);
    for (j = 0; j < mlen; j++)
        m[j] = mod[j + 1];

    /* Shift m left to make msb bit set */
    for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
        if ((m[mlen - 1] << mshift) & BIGNUM_TOP_BIT)
            break;
    if (mshift)
        shift_left(m, mlen, mshift);

    /* Allocate n of size mlen, copy base to n */
    n = snewn(mlen, BignumInt);
    for (i = 0; i < (int)base[0]; i++)
        n[i] = base[i + 1];
    for (; i < mlen; i++)
        n[i] = 0;

    /* Allocate a and b of size 2*mlen. Set a = 1 */
    a = snewn(2 * mlen, BignumInt);
    b = snewn(2 * mlen, BignumInt);
    a[0] = 1;
    for (i = 1; i < 2 * mlen; i++)
        a[i] = 0;

    /* Scratch space for multiplies */
    scratchlen = mul_compute_scratch(mlen);
    scratch = snewn(scratchlen, BignumInt);

    /* Skip leading zero bits of exp. */
    i = 0;
    j = BIGNUM_INT_BITS-1;
    while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
        j--;
        if (j < 0) {
            i++;
            j = BIGNUM_INT_BITS-1;
        }
    }

    /* Main computation */
    while (i < (int)exp[0]) {
        while (j >= 0) {
            internal_mul(a, a, b, mlen, scratch);
            internal_mod(b, mlen * 2, m, mlen, NULL, 0);
            if ((exp[exp[0] - i] & (1 << j)) != 0) {
                internal_mul(b, n, a, mlen, scratch);
                internal_mod(a, mlen * 2, m, mlen, NULL, 0);
            } else {
                BignumInt *t;
                t = a;
                a = b;
                b = t;
            }
            j--;
        }
        i++;
        j = BIGNUM_INT_BITS-1;
    }

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
        shift_left(a, mlen + 1, mshift);
        internal_mod(a, mlen + 1, m, mlen, NULL, 0);
        shift_right(a, mlen, mshift);
    }

    /* Copy result to buffer */
    result = newbn(mod[0]);
    for (i = 0; i < mlen; i++)
        result[i + 1] = a[i];
    while (result[0] > 1 && result[result[0]] == 0)
        result[0]--;

    /* Free temporary arrays */
    for (i = 0; i < 2 * mlen; i++)
        a[i] = 0;
    sfree(a);
    for (i = 0; i < scratchlen; i++)
        scratch[i] = 0;
    sfree(scratch);
    for (i = 0; i < 2 * mlen; i++)
        b[i] = 0;
    sfree(b);
    for (i = 0; i < mlen; i++)
        m[i] = 0;
    sfree(m);
    for (i = 0; i < mlen; i++)
        n[i] = 0;
    sfree(n);

    freebn(base);

    return result;
}

/*
 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
 * technique where possible, falling back to modpow_simple otherwise.
 */
Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
{
    BignumInt *a, *b, *x, *n, *mninv, *scratch;
    int len, scratchlen, i, j;
    Bignum base, base2, r, rn, inv, result;

    /*
     * The most significant word of mod needs to be non-zero. It
     * should already be, but let's make sure.
     */
    assert(mod[mod[0]] != 0);

    /*
     * mod had better be odd, or we can't do Montgomery multiplication
     * using a power of two at all.
     */
    if (!(mod[1] & 1))
        return modpow_simple(base_in, exp, mod);

    /*
     * Make sure the base is smaller than the modulus, by reducing
     * it modulo the modulus if not.
     */
    base = bigmod(base_in, mod);

    /*
     * Compute the inverse of n mod r, for monty_reduce. (In fact we
     * want the inverse of _minus_ n mod r, but we'll sort that out
     * below.)
     */
    len = mod[0];
    r = bn_power_2(BIGNUM_INT_BITS * len);
    inv = modinv(mod, r);

    /*
     * Multiply the base by r mod n, to get it into Montgomery
     * representation.
     */
    base2 = modmul(base, r, mod);
    freebn(base);
    base = base2;

    rn = bigmod(r, mod);               /* r mod n, i.e. Montgomerified 1 */

    freebn(r);                         /* won't need this any more */

    /*
     * Set up internal arrays of the right lengths containing the base,
     * the modulus, and the modulus's inverse.
     */
    n = snewn(len, BignumInt);
    for (j = 0; j < len; j++)
        n[j] = mod[j + 1];

    mninv = snewn(len, BignumInt);
    for (j = 0; j < len; j++)
        mninv[j] = (j < (int)inv[0] ? inv[j + 1] : 0);
    freebn(inv);         /* we don't need this copy of it any more */
    /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
    x = snewn(len, BignumInt);
    for (j = 0; j < len; j++)
        x[j] = 0;
    internal_sub(x, mninv, mninv, len);

    /* x = snewn(len, BignumInt); */ /* already done above */
    for (j = 0; j < len; j++)
        x[j] = (j < (int)base[0] ? base[j + 1] : 0);
    freebn(base);        /* we don't need this copy of it any more */

    a = snewn(2*len, BignumInt);
    b = snewn(2*len, BignumInt);
    for (j = 0; j < len; j++)
        a[j] = (j < (int)rn[0] ? rn[j + 1] : 0);
    freebn(rn);

    /* Scratch space for multiplies */
    scratchlen = 3*len + mul_compute_scratch(len);
    scratch = snewn(scratchlen, BignumInt);

    /* Skip leading zero bits of exp. */
    i = 0;
    j = BIGNUM_INT_BITS-1;
    while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
        j--;
        if (j < 0) {
            i++;
            j = BIGNUM_INT_BITS-1;
        }
    }

    /* Main computation */
    while (i < (int)exp[0]) {
        while (j >= 0) {
            internal_mul(a, a, b, len, scratch);
            monty_reduce(b, n, mninv, scratch, len);
            if ((exp[exp[0] - i] & (1 << j)) != 0) {
                internal_mul(b, x, a, len,  scratch);
                monty_reduce(a, n, mninv, scratch, len);
            } else {
                BignumInt *t;
                t = a;
                a = b;
                b = t;
            }
            j--;
        }
        i++;
        j = BIGNUM_INT_BITS-1;
    }

    /*
     * Final monty_reduce to get back from the adjusted Montgomery
     * representation.
     */
    monty_reduce(a, n, mninv, scratch, len);

    /* Copy result to buffer */
    result = newbn(mod[0]);
    for (i = 0; i < len; i++)
        result[i + 1] = a[i];
    while (result[0] > 1 && result[result[0]] == 0)
        result[0]--;

    /* Free temporary arrays */
    for (i = 0; i < scratchlen; i++)
        scratch[i] = 0;
    sfree(scratch);
    for (i = 0; i < 2 * len; i++)
        a[i] = 0;
    sfree(a);
    for (i = 0; i < 2 * len; i++)
        b[i] = 0;
    sfree(b);
    for (i = 0; i < len; i++)
        mninv[i] = 0;
    sfree(mninv);
    for (i = 0; i < len; i++)
        n[i] = 0;
    sfree(n);
    for (i = 0; i < len; i++)
        x[i] = 0;
    sfree(x);

    return result;
}

/*
 * Compute (p * q) % mod.
 * The most significant word of mod MUST be non-zero.
 * We assume that the result array is the same size as the mod array.
 */
Bignum modmul(Bignum p, Bignum q, Bignum mod)
{
    BignumInt *a, *n, *m, *o, *scratch;
    int mshift, scratchlen;
    int pqlen, mlen, rlen, i, j;
    Bignum result;

    /* Allocate m of size mlen, copy mod to m */
    mlen = mod[0];
    m = snewn(mlen, BignumInt);
    for (j = 0; j < mlen; j++)
        m[j] = mod[j + 1];

    /* Shift m left to make msb bit set */
    for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
        if ((m[mlen - 1] << mshift) & BIGNUM_TOP_BIT)
            break;
    if (mshift)
        shift_left(m, mlen, mshift);

    pqlen = (p[0] > q[0] ? p[0] : q[0]);

    /* Make sure that we're allowing enough space.  The shifting below will
     * underflow the vectors we allocate if `pqlen' is too small.
     */
    if (2*pqlen <= mlen)
        pqlen = mlen/2 + 1;

    /* Allocate n of size pqlen, copy p to n */
    n = snewn(pqlen, BignumInt);
    for (i = 0; i < (int)p[0]; i++)
        n[i] = p[i + 1];
    for (; i < pqlen; i++)
        n[i] = 0;

    /* Allocate o of size pqlen, copy q to o */
    o = snewn(pqlen, BignumInt);
    for (i = 0; i < (int)q[0]; i++)
        o[i] = q[i + 1];
    for (; i < pqlen; i++)
        o[i] = 0;

    /* Allocate a of size 2*pqlen for result */
    a = snewn(2 * pqlen, BignumInt);

    /* Scratch space for multiplies */
    scratchlen = mul_compute_scratch(pqlen);
    scratch = snewn(scratchlen, BignumInt);

    /* Main computation */
    internal_mul(n, o, a, pqlen, scratch);
    internal_mod(a, pqlen * 2, m, mlen, NULL, 0);

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
        shift_left(a, mlen + 1, mshift);
        internal_mod(a, mlen + 1, m, mlen, NULL, 0);
        shift_right(a, mlen, mshift);
    }

    /* Copy result to buffer */
    rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
    result = newbn(rlen);
    for (i = 0; i < rlen; i++)
        result[i + 1] = a[i];
    while (result[0] > 1 && result[result[0]] == 0)
        result[0]--;

    /* Free temporary arrays */
    for (i = 0; i < scratchlen; i++)
        scratch[i] = 0;
    sfree(scratch);
    for (i = 0; i < 2 * pqlen; i++)
        a[i] = 0;
    sfree(a);
    for (i = 0; i < mlen; i++)
        m[i] = 0;
    sfree(m);
    for (i = 0; i < pqlen; i++)
        n[i] = 0;
    sfree(n);
    for (i = 0; i < pqlen; i++)
        o[i] = 0;
    sfree(o);

    return result;
}

/*
 * Compute p % mod.
 * The most significant word of mod MUST be non-zero.
 * We assume that the result array is the same size as the mod array.
 * We optionally write out a quotient if `quotient' is non-NULL.
 * We can avoid writing out the result if `result' is NULL.
 */
static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
{
    BignumInt *n, *m;
    int mshift;
    int plen, mlen, i, j;

    /* Allocate m of size mlen, copy mod to m */
    mlen = mod[0];
    m = snewn(mlen, BignumInt);
    for (j = 0; j < mlen; j++)
        m[j] = mod[j + 1];

    /* Shift m left to make msb bit set */
    for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
        if ((m[mlen - 1] << mshift) & BIGNUM_TOP_BIT)
            break;
    if (mshift)
        shift_left(m, mlen, mshift);

    plen = p[0];
    /* Ensure plen > mlen */
    if (plen <= mlen)
        plen = mlen + 1;

    /* Allocate n of size plen, copy p to n */
    n = snewn(plen, BignumInt);
    for (i = 0; i < (int)p[0]; i++)
        n[i] = p[i + 1];
    for (; i < plen; i++)
        n[i] = 0;

    /* Main computation */
    internal_mod(n, plen, m, mlen, quotient, mshift);

    /* Fixup result in case the modulus was shifted */
    if (mshift) {
        shift_left(n, mlen + 1, mshift);
        internal_mod(n, plen, m, mlen, quotient, 0);
        shift_right(n, mlen, mshift);
    }

    /* Copy result to buffer */
    if (result) {
        for (i = 0; i < (int)result[0]; i++)
            result[i + 1] = i < plen ? n[i] : 0;
        bn_restore_invariant(result);
    }

    /* Free temporary arrays */
    for (i = 0; i < mlen; i++)
        m[i] = 0;
    sfree(m);
    for (i = 0; i < plen; i++)
        n[i] = 0;
    sfree(n);
}

/*
 * Decrement a number.
 */
void decbn(Bignum bn)
{
    int i = 1;
    while (i < (int)bn[0] && bn[i] == 0)
        bn[i++] = BIGNUM_INT_MASK;
    bn[i]--;
}

Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
{
    Bignum result;
    int w, i;

    w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */

    result = newbn(w);
    for (i = 1; i <= w; i++)
        result[i] = 0;
    for (i = nbytes; i--;) {
        unsigned char byte = *data++;
        result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
    }

    while (result[0] > 1 && result[result[0]] == 0)
        result[0]--;
    return result;
}

/*
 * Read an SSH-1-format bignum from a data buffer. Return the number
 * of bytes consumed, or -1 if there wasn't enough data.
 */
int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
{
    const unsigned char *p = data;
    int i;
    int w, b;

    if (len < 2)
        return -1;

    w = 0;
    for (i = 0; i < 2; i++)
        w = (w << 8) + *p++;
    b = (w + 7) / 8;                   /* bits -> bytes */

    if (len < b+2)
        return -1;

    if (!result)                       /* just return length */
        return b + 2;

    *result = bignum_from_bytes(p, b);

    return p + b - data;
}

/*
 * Return the bit count of a bignum, for SSH-1 encoding.
 */
int bignum_bitcount(Bignum bn)
{
    int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
    while (bitcount >= 0
           && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
    return bitcount + 1;
}

/*
 * Return the byte length of a bignum when SSH-1 encoded.
 */
int ssh1_bignum_length(Bignum bn)
{
    return 2 + (bignum_bitcount(bn) + 7) / 8;
}

/*
 * Return the byte length of a bignum when SSH-2 encoded.
 */
int ssh2_bignum_length(Bignum bn)
{
    return 4 + (bignum_bitcount(bn) + 8) / 8;
}

/*
 * Return a byte from a bignum; 0 is least significant, etc.
 */
int bignum_byte(Bignum bn, int i)
{
    if (i >= (int)(BIGNUM_INT_BYTES * bn[0]))
        return 0;                      /* beyond the end */
    else
        return (bn[i / BIGNUM_INT_BYTES + 1] >>
                ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
}

/*
 * Return a bit from a bignum; 0 is least significant, etc.
 */
int bignum_bit(Bignum bn, int i)
{
    if (i >= (int)(BIGNUM_INT_BITS * bn[0]))
        return 0;                      /* beyond the end */
    else
        return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
}

/*
 * Set a bit in a bignum; 0 is least significant, etc.
 */
void bignum_set_bit(Bignum bn, int bitnum, int value)
{
    if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
        abort();                       /* beyond the end */
    else {
        int v = bitnum / BIGNUM_INT_BITS + 1;
        int mask = 1 << (bitnum % BIGNUM_INT_BITS);
        if (value)
            bn[v] |= mask;
        else
            bn[v] &= ~mask;
    }
}

/*
 * Write a SSH-1-format bignum into a buffer. It is assumed the
 * buffer is big enough. Returns the number of bytes used.
 */
int ssh1_write_bignum(void *data, Bignum bn)
{
    unsigned char *p = data;
    int len = ssh1_bignum_length(bn);
    int i;
    int bitc = bignum_bitcount(bn);

    *p++ = (bitc >> 8) & 0xFF;
    *p++ = (bitc) & 0xFF;
    for (i = len - 2; i--;)
        *p++ = bignum_byte(bn, i);
    return len;
}

/*
 * Compare two bignums. Returns like strcmp.
 */
int bignum_cmp(Bignum a, Bignum b)
{
    int amax = a[0], bmax = b[0];
    int i = (amax > bmax ? amax : bmax);
    while (i) {
        BignumInt aval = (i > amax ? 0 : a[i]);
        BignumInt bval = (i > bmax ? 0 : b[i]);
        if (aval < bval)
            return -1;
        if (aval > bval)
            return +1;
        i--;
    }
    return 0;
}

/*
 * Right-shift one bignum to form another.
 */
Bignum bignum_rshift(Bignum a, int shift)
{
    Bignum ret;
    int i, shiftw, shiftb, shiftbb, bits;
    BignumInt ai, ai1;

    bits = bignum_bitcount(a) - shift;
    ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);

    if (ret) {
        shiftw = shift / BIGNUM_INT_BITS;
        shiftb = shift % BIGNUM_INT_BITS;
        shiftbb = BIGNUM_INT_BITS - shiftb;

        ai1 = a[shiftw + 1];
        for (i = 1; i <= (int)ret[0]; i++) {
            ai = ai1;
            ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
            ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
        }
    }

    return ret;
}

/*
 * Non-modular multiplication and addition.
 */
Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
{
    int alen = a[0], blen = b[0];
    int mlen = (alen > blen ? alen : blen);
    int rlen, i, maxspot;
    int wslen;
    BignumInt *workspace;
    Bignum ret;

    /* mlen space for a, mlen space for b, 2*mlen for result,
     * plus scratch space for multiplication */
    wslen = mlen * 4 + mul_compute_scratch(mlen);
    workspace = snewn(wslen, BignumInt);
    for (i = 0; i < mlen; i++) {
        workspace[0 * mlen + i] = i < (int)a[0] ? a[i + 1] : 0;
        workspace[1 * mlen + i] = i < (int)b[0] ? b[i + 1] : 0;
    }

    internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
                 workspace + 2 * mlen, mlen, workspace + 4 * mlen);

    /* now just copy the result back */
    rlen = alen + blen + 1;
    if (addend && rlen <= (int)addend[0])
        rlen = addend[0] + 1;
    ret = newbn(rlen);
    maxspot = 0;
    for (i = 0; i < (int)ret[0]; i++) {
        ret[i + 1] = (i < 2 * mlen ? workspace[2 * mlen + i] : 0);
        if (ret[i + 1] != 0)
            maxspot = i + 1;
    }
    ret[0] = maxspot;

    /* now add in the addend, if any */
    if (addend) {
        BignumDblInt carry = 0;
        for (i = 1; i <= rlen; i++) {
            carry += (i <= (int)ret[0] ? ret[i] : 0);
            carry += (i <= (int)addend[0] ? addend[i] : 0);
            ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
            carry >>= BIGNUM_INT_BITS;
            if (ret[i] != 0 && i > maxspot)
                maxspot = i;
        }
    }
    ret[0] = maxspot;

    for (i = 0; i < wslen; i++)
        workspace[i] = 0;
    sfree(workspace);
    return ret;
}

/*
 * Non-modular multiplication.
 */
Bignum bigmul(Bignum a, Bignum b)
{
    return bigmuladd(a, b, NULL);
}

/*
 * Simple addition.
 */
Bignum bigadd(Bignum a, Bignum b)
{
    int alen = a[0], blen = b[0];
    int rlen = (alen > blen ? alen : blen) + 1;
    int i, maxspot;
    Bignum ret;
    BignumDblInt carry;

    ret = newbn(rlen);

    carry = 0;
    maxspot = 0;
    for (i = 1; i <= rlen; i++) {
        carry += (i <= (int)a[0] ? a[i] : 0);
        carry += (i <= (int)b[0] ? b[i] : 0);
        ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
        carry >>= BIGNUM_INT_BITS;
        if (ret[i] != 0 && i > maxspot)
            maxspot = i;
    }
    ret[0] = maxspot;

    return ret;
}

/*
 * Subtraction. Returns a-b, or NULL if the result would come out
 * negative (recall that this entire bignum module only handles
 * positive numbers).
 */
Bignum bigsub(Bignum a, Bignum b)
{
    int alen = a[0], blen = b[0];
    int rlen = (alen > blen ? alen : blen);
    int i, maxspot;
    Bignum ret;
    BignumDblInt carry;

    ret = newbn(rlen);

    carry = 1;
    maxspot = 0;
    for (i = 1; i <= rlen; i++) {
        carry += (i <= (int)a[0] ? a[i] : 0);
        carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
        ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
        carry >>= BIGNUM_INT_BITS;
        if (ret[i] != 0 && i > maxspot)
            maxspot = i;
    }
    ret[0] = maxspot;

    if (!carry) {
        freebn(ret);
        return NULL;
    }

    return ret;
}

/*
 * Return a bignum which is the result of shifting another left by N bits.
 * If N is negative then you get a right shift instead.
 */
Bignum biglsl(Bignum x, int n)
{
    Bignum d;
    unsigned o, i;

    if (!n || !x[0])
        return copybn(x);
    else if (n < 0)
        return biglsr(x, -n);

    o = n/BIGNUM_INT_BITS;
    n %= BIGNUM_INT_BITS;
    d = newbn(x[0] + o + !!n);

    for (i = 1; i <= o; i++)
        d[i] = 0;

    if (!n) {
        for (i = 1; i <= x[0]; i++)
            d[o + i] = x[i];
    } else {
        d[o + 1] = x[1] << n;
        for (i = 2; i <= x[0]; i--)
            d[o + i] = (x[i] << n) | (x[i - 1] >> (BIGNUM_INT_BITS - n));
        d[o + x[0] + 1] = x[x[0]] >> (BIGNUM_INT_BITS - n);
    }

    bn_restore_invariant(d);
    return d;
}

/*
 * Return a bignum which is the result of shifting another right by N bits
 * (discarding the least significant N bits, and shifting zeroes in at the
 * most significant end).  If N is negative then you get a left shift
 * instead.
 */
Bignum biglsr(Bignum x, int n)
{
    Bignum d;
    unsigned o, i;

    if (!n || !x[0])
        return copybn(x);
    else if (n < 0)
        return biglsl(x, -n);

    o = n/BIGNUM_INT_BITS;
    n %= BIGNUM_INT_BITS;
    d = newbn(x[0]);

    if (!n) {
        for (i = o + 1; i <= x[0]; i++)
            d[i - o] = x[i];
    } else {
        d[1] = x[o + 1] >> n;
        for (i = o + 2; i < x[0]; i++)
            d[i - o] = x[
        d[o + x[0] + 1] = x[x[0]] >> (BIGNUM_INT_BITS - n);
        for (i = x[0]; i > 1; i--)
            d[o + i] = (x[i] << n) | (x[i - 1] >> (BIGNUM_INT_BITS - n));
        d[o + 1] = x[1] << n;
    }

    bn_restore_invariant(d);
    return d;
}

/*
 * Create a bignum which is the bitmask covering another one. That
 * is, the smallest integer which is >= N and is also one less than
 * a power of two.
 */
Bignum bignum_bitmask(Bignum n)
{
    Bignum ret = copybn(n);
    int i;
    BignumInt j;

    i = ret[0];
    while (n[i] == 0 && i > 0)
        i--;
    if (i <= 0)
        return ret;                    /* input was zero */
    j = 1;
    while (j < n[i])
        j = 2 * j + 1;
    ret[i] = j;
    while (--i > 0)
        ret[i] = BIGNUM_INT_MASK;
    return ret;
}

/*
 * Convert a (max 32-bit) long into a bignum.
 */
Bignum bignum_from_long(unsigned long nn)
{
    Bignum ret;
    BignumDblInt n = nn;

    ret = newbn(3);
    ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
    ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
    ret[3] = 0;
    ret[0] = (ret[2]  ? 2 : 1);
    return ret;
}

/*
 * Add a long to a bignum.
 */
Bignum bignum_add_long(Bignum number, unsigned long addendx)
{
    Bignum ret = newbn(number[0] + 1);
    int i, maxspot = 0;
    BignumDblInt carry = 0, addend = addendx;

    for (i = 1; i <= (int)ret[0]; i++) {
        carry += addend & BIGNUM_INT_MASK;
        carry += (i <= (int)number[0] ? number[i] : 0);
        addend >>= BIGNUM_INT_BITS;
        ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
        carry >>= BIGNUM_INT_BITS;
        if (ret[i] != 0)
            maxspot = i;
    }
    ret[0] = maxspot;
    return ret;
}

/*
 * Compute the residue of a bignum, modulo a (max 16-bit) short.
 */
unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
{
    BignumDblInt mod, r;
    int i;

    r = 0;
    mod = modulus;
    for (i = number[0]; i > 0; i--)
        r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
    return (unsigned short) r;
}

#ifdef DEBUG
void diagbn(char *prefix, Bignum md)
{
    int i, nibbles, morenibbles;
    static const char hex[] = "0123456789ABCDEF";

    debug(("%s0x", prefix ? prefix : ""));

    nibbles = (3 + bignum_bitcount(md)) / 4;
    if (nibbles < 1)
        nibbles = 1;
    morenibbles = 4 * md[0] - nibbles;
    for (i = 0; i < morenibbles; i++)
        debug(("-"));
    for (i = nibbles; i--;)
        debug(("%c",
               hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));

    if (prefix)
        debug(("\n"));
}
#endif

/*
 * Simple division.
 */
Bignum bigdiv(Bignum a, Bignum b)
{
    Bignum q = newbn(a[0]);
    bigdivmod(a, b, NULL, q);
    return q;
}

/*
 * Simple remainder.
 */
Bignum bigmod(Bignum a, Bignum b)
{
    Bignum r = newbn(b[0]);
    bigdivmod(a, b, r, NULL);
    return r;
}

/*
 * Greatest common divisor.
 */
Bignum biggcd(Bignum av, Bignum bv)
{
    Bignum a = copybn(av);
    Bignum b = copybn(bv);

    while (bignum_cmp(b, Zero) != 0) {
        Bignum t = newbn(b[0]);
        bigdivmod(a, b, t, NULL);
        while (t[0] > 1 && t[t[0]] == 0)
            t[0]--;
        freebn(a);
        a = b;
        b = t;
    }

    freebn(b);
    return a;
}

/*
 * Modular inverse, using Euclid's extended algorithm.
 */
Bignum modinv(Bignum number, Bignum modulus)
{
    Bignum a = copybn(modulus);
    Bignum b = copybn(number);
    Bignum xp = copybn(Zero);
    Bignum x = copybn(One);
    int sign = +1;

    while (bignum_cmp(b, One) != 0) {
        Bignum t = newbn(b[0]);
        Bignum q = newbn(a[0]);
        bigdivmod(a, b, t, q);
        while (t[0] > 1 && t[t[0]] == 0)
            t[0]--;
        freebn(a);
        a = b;
        b = t;
        t = xp;
        xp = x;
        x = bigmuladd(q, xp, t);
        sign = -sign;
        freebn(t);
        freebn(q);
    }

    freebn(b);
    freebn(a);
    freebn(xp);

    /* now we know that sign * x == 1, and that x < modulus */
    if (sign < 0) {
        /* set a new x to be modulus - x */
        Bignum newx = newbn(modulus[0]);
        BignumInt carry = 0;
        int maxspot = 1;
        int i;

        for (i = 1; i <= (int)newx[0]; i++) {
            BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
            BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
            newx[i] = aword - bword - carry;
            bword = ~bword;
            carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
            if (newx[i] != 0)
                maxspot = i;
        }
        newx[0] = maxspot;
        freebn(x);
        x = newx;
    }

    /* and return. */
    return x;
}

/*
 * Extract the largest power of 2 dividing x, storing it in p2, and returning
 * the product of the remaining factors.
 */
static Bignum extract_p2(Bignum x, unsigned *p2)
{
    unsigned i, j, k, n;
    Bignum y;

    /* If x is zero then the following won't work.  And if x is odd then
     * there's nothing very useful to do.
     */
    if (!x[0] || (x[1] & 1)) {
        *p2 = 0;
        return copybn(x);
    }

    /* Find the power of two. */
    for (i = 0; !x[i + 1]; i++);
    for (j = 0; !((x[i + 1] >> j) & 1); j++);
    *p2 = i*BIGNUM_INT_BITS + j;

    /* Work out how big the copy should be. */
    n = x[0] - i - 1;
    if (x[x[0]] >> j) n++;

    /* Copy and shift down. */
    y = newbn(n);
    for (k = 1; k <= n; k++) {
        y[k] = x[k + i] >> j;
        if (j && k < x[0]) y[k] |= x[k + i + 1] << (BIGNUM_INT_BITS - j);
    }

    /* Done. */
    return y;
}

/*
 * Kronecker symbol (a|n).  The result is always in { -1, 0, +1 }, and is
 * zero if and only if a and n have a nontrivial common factor.  Most
 * usefully, if n is prime, this is the Legendre symbol, taking the value +1
 * if a is a quadratic residue mod n, and -1 otherwise; i.e., (a|p) ==
 * a^{(p-1)/2} (mod p).
 */
int kronecker(Bignum a, Bignum n)
{
    unsigned s, nn;
    int r = +1;
    Bignum t;

    /* Special case for n = 0.  This is the same convention PARI uses,
     * except that we can't represent negative numbers.
     */
    if (bignum_cmp(n, Zero) == 0) {
        if (bignum_cmp(a, One) == 0) return +1;
        else return 0;
    }

    /* Write n = 2^s t, with t odd.  If s > 0 and a is even, then the answer
     * is zero; otherwise throw in a factor of (-1)^s if a == 3 or 5 (mod 8).
     *
     * At this point, we have a copy of n, and must remember to free it when
     * we're done.  It's convenient to take a copy of a at the same time.
     */
    a = copybn(a);
    n = extract_p2(n, &s);

    if (s && (!a[0] || !(a[1] & 1))) { r = 0; goto done; }
    else if ((s & 1) && ((a[1] & 7) == 3 || (a[1] & 7) == 5)) r = -r;

    /* If n is (now) a unit then we're done. */
    if (bignum_cmp(n, One) == 0) goto done;

    /* Reduce a modulo n before we go any further. */
    if (bignum_cmp(a, n) >= 0) { t = bigmod(a, n); freebn(a); a = t; }

    /* Main loop. */
    for (;;) {
        if (bignum_cmp(a, Zero) == 0) { r = 0; goto done; }

        /* Strip out and handle powers of two from a. */
        t = extract_p2(a, &s); freebn(a); a = t;
        nn = n[1] & 7;
        if ((s & 1) && (nn == 3 || nn == 5)) r = -r;
        if (bignum_cmp(a, One) == 0) break;

        /* Swap, applying quadratic reciprocity. */
        if ((nn & 3) == 3 && (a[1] & 3) == 3) r = -r;
        t = bigmod(n, a); freebn(n); n = a; a = t;
    }

    /* Tidy up: we're done. */
done:
    freebn(a); freebn(n);
    return r;
}

/*
 * Modular square root.  We must have p prime: extracting square roots modulo
 * composites is equivalent to factoring (but we don't check: you'll just get
 * the wrong answer).  Returns NULL if x is not a quadratic residue mod p.
 */
Bignum modsqrt(Bignum x, Bignum p)
{
    Bignum xinv, b, c, r, t, z, X, mone;
    unsigned i, j, s;

    /* If x is not a quadratic residue then we will not go to space today. */
    if (kronecker(x, p) != +1) return NULL;

    /* We need a quadratic nonresidue from somewhere.  Exactly half of all
     * units mod p are quadratic residues, but no efficient deterministic
     * algorithm for finding one is known.  So pick at random: we don't
     * expect this to take long.
     */
    z = newbn(p[0]);
    do {
        for (i = 1; i <= p[0]; i++) z[i] = rand();
        z[0] = p[0]; bn_restore_invariant(z);
    } while (kronecker(z, p) != -1);
    b = bigmod(z, p); freebn(z);

    /* We need to compute a few things before we really get started. */
    xinv = modinv(x, p);                /* x^{-1} mod p */
    mone = bigsub(p, One);              /* p - 1 == -1 (mod p) */
    t = extract_p2(mone, &s);           /* 2^s t = p - 1 */
    c = modpow(b, t, p);                /* b^t (mod p) */
    z = bigadd(t, One); freebn(t); t = z; /* (t + 1) */
    shift_right(t + 1, t[0], 1); if (!t[t[0]]) t[0]--;
    r = modpow(x, t, p);                /* x^{(t+1)/2} (mod p) */
    freebn(b); freebn(mone); freebn(t);

    /* OK, so how does this work anyway?
     *
     * We know that x^t is somewhere in the order-2^s subgroup of GF(p)^*;
     * and g = c^{-1} is a generator for this subgroup (since we know that
     * g^{2^{s-1}} = b^{(p-1)/2} = (b|p) = -1); so x^t = g^m for some m.  In
     * fact, we know that m is even because x is a square.  Suppose we can
     * determine m; then we know that x^t/g^m = 1, so x^{t+1}/c^m = x -- but
     * both t + 1 and m are even, so x^{(t+1)/2}/g^{m/2} is a square root of
     * x.
     *
     * Conveniently, finding the discrete log of an element X in a group of
     * order 2^s is easy.  Write X = g^m = g^{m_0+2k'}; then X^{2^{s-1}} =
     * g^{m_0 2^{s-1}} c^{m' 2^s} = g^{m_0 2^{s-1}} is either -1 or +1,
     * telling us that m_0 is 1 or 0 respectively.  Then X/g^{m_0} =
     * (g^2)^{m'} has order 2^{s-1} so we can continue inductively.  What we
     * end up with at the end is X/g^m.
     *
     * There are a few wrinkles.  As we proceed through the induction, the
     * generator for the subgroup will be c^{-2}, since we know that m is
     * even.  While we want the discrete log of X = x^t, we're actually going
     * to keep track of r, which will eventually be x^{(t+1)/2}/g^{m/2} =
     * x^{(t+1)/2} c^m, recovering X/g^m = r^2/x as we go.  We don't actually
     * form the discrete log explicitly, because the final result will
     * actually be the square root we want.
     */
    for (i = 1; i < s; i++) {

        /* Determine X.  We could optimize this, only recomputing it when
         * it's been invalidated, but that's fiddlier and this isn't
         * performance critical.
         */
        z = modmul(r, r, p);
        X = modmul(z, xinv, p);
        freebn(z);

        /* Determine X^{2^{s-1-i}}. */
        for (j = i + 1; j < s; j++)
            z = modmul(X, X, p), freebn(X), X = z;

        /* Maybe accumulate a factor of c. */
        if (bignum_cmp(X, One) != 0)
            z = modmul(r, c, p), freebn(r), r = z;

        /* Move on to the next smaller subgroup. */
        z = modmul(c, c, p), freebn(c), c = z;
        freebn(X);
    }

    /* Of course, there are two square roots of x.  For predictability's sake
     * we'll always return the one in [1..(p - 1)/2].  The other is, of
     * course, p - r.
     */
    z = bigsub(p, r);
    if (bignum_cmp(r, z) < 0)
        freebn(z);
    else {
        freebn(r);
        r = z;
    }

    /* We're done. */
    freebn(xinv); freebn(c);
    return r;
}

/*
 * Render a bignum into decimal. Return a malloced string holding
 * the decimal representation.
 */
char *bignum_decimal(Bignum x)
{
    int ndigits, ndigit;
    int i, iszero;
    BignumDblInt carry;
    char *ret;
    BignumInt *workspace;

    /*
     * First, estimate the number of digits. Since log(10)/log(2)
     * is just greater than 93/28 (the joys of continued fraction
     * approximations...) we know that for every 93 bits, we need
     * at most 28 digits. This will tell us how much to malloc.
     *
     * Formally: if x has i bits, that means x is strictly less
     * than 2^i. Since 2 is less than 10^(28/93), this is less than
     * 10^(28i/93). We need an integer power of ten, so we must
     * round up (rounding down might make it less than x again).
     * Therefore if we multiply the bit count by 28/93, rounding
     * up, we will have enough digits.
     *
     * i=0 (i.e., x=0) is an irritating special case.
     */
    i = bignum_bitcount(x);
    if (!i)
        ndigits = 1;                   /* x = 0 */
    else
        ndigits = (28 * i + 92) / 93;  /* multiply by 28/93 and round up */
    ndigits++;                         /* allow for trailing \0 */
    ret = snewn(ndigits, char);

    /*
     * Now allocate some workspace to hold the binary form as we
     * repeatedly divide it by ten. Initialise this to the
     * big-endian form of the number.
     */
    workspace = snewn(x[0], BignumInt);
    for (i = 0; i < (int)x[0]; i++)
        workspace[i] = x[x[0] - i];

    /*
     * Next, write the decimal number starting with the last digit.
     * We use ordinary short division, dividing 10 into the
     * workspace.
     */
    ndigit = ndigits - 1;
    ret[ndigit] = '\0';
    do {
        iszero = 1;
        carry = 0;
        for (i = 0; i < (int)x[0]; i++) {
            carry = (carry << BIGNUM_INT_BITS) + workspace[i];
            workspace[i] = (BignumInt) (carry / 10);
            if (workspace[i])
                iszero = 0;
            carry %= 10;
        }
        ret[--ndigit] = (char) (carry + '0');
    } while (!iszero);

    /*
     * There's a chance we've fallen short of the start of the
     * string. Correct if so.
     */
    if (ndigit > 0)
        memmove(ret, ret + ndigit, ndigits - ndigit);

    /*
     * Done.
     */
    sfree(workspace);
    return ret;
}

#ifdef TESTBN

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>

/*
 * gcc -Wall -g -O0 -DTESTBN -o testbn sshbn.c misc.c conf.c tree234.c unix/uxmisc.c -I. -I unix -I charset
 *
 * Then feed to this program's standard input the output of
 * testdata/bignum.py .
 */

void modalfatalbox(char *p, ...)
{
    va_list ap;
    fprintf(stderr, "FATAL ERROR: ");
    va_start(ap, p);
    vfprintf(stderr, p, ap);
    va_end(ap);
    fputc('\n', stderr);
    exit(1);
}

#define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )

int main(int argc, char **argv)
{
    char *buf;
    int line = 0;
    int passes = 0, fails = 0;

    while ((buf = fgetline(stdin)) != NULL) {
        int maxlen = strlen(buf);
        unsigned char *data = snewn(maxlen, unsigned char);
        unsigned char *ptrs[5], *q;
        int ptrnum;
        char *bufp = buf;

        line++;

        q = data;
        ptrnum = 0;

        while (*bufp && !isspace((unsigned char)*bufp))
            bufp++;
        if (bufp)
            *bufp++ = '\0';

        while (*bufp) {
            char *start, *end;
            int i;

            while (*bufp && !isxdigit((unsigned char)*bufp))
                bufp++;
            start = bufp;

            if (!*bufp)
                break;

            while (*bufp && isxdigit((unsigned char)*bufp))
                bufp++;
            end = bufp;

            if (ptrnum >= lenof(ptrs))
                break;
            ptrs[ptrnum++] = q;
            
            for (i = -((end - start) & 1); i < end-start; i += 2) {
                unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
                val = val * 16 + fromxdigit(start[i+1]);
                *q++ = val;
            }

            ptrs[ptrnum] = q;
        }

        if (!strcmp(buf, "mul")) {
            Bignum a, b, c, p;

            if (ptrnum != 3) {
                printf("%d: mul with %d parameters, expected 3\n", line, ptrnum);
                exit(1);
            }
            a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
            b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
            c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
            p = bigmul(a, b);

            if (bignum_cmp(c, p) == 0) {
                passes++;
            } else {
                char *as = bignum_decimal(a);
                char *bs = bignum_decimal(b);
                char *cs = bignum_decimal(c);
                char *ps = bignum_decimal(p);
                
                printf("%d: fail: %s * %s gave %s expected %s\n",
                       line, as, bs, ps, cs);
                fails++;

                sfree(as);
                sfree(bs);
                sfree(cs);
                sfree(ps);
            }
            freebn(a);
            freebn(b);
            freebn(c);
            freebn(p);
        } else if (!strcmp(buf, "pow")) {
            Bignum base, expt, modulus, expected, answer;

            if (ptrnum != 4) {
                printf("%d: mul with %d parameters, expected 4\n", line, ptrnum);
                exit(1);
            }

            base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
            expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
            modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
            expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
            answer = modpow(base, expt, modulus);

            if (bignum_cmp(expected, answer) == 0) {
                passes++;
            } else {
                char *as = bignum_decimal(base);
                char *bs = bignum_decimal(expt);
                char *cs = bignum_decimal(modulus);
                char *ds = bignum_decimal(answer);
                char *ps = bignum_decimal(expected);
                
                printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
                       line, as, bs, cs, ds, ps);
                fails++;

                sfree(as);
                sfree(bs);
                sfree(cs);
                sfree(ds);
                sfree(ps);
            }
            freebn(base);
            freebn(expt);
            freebn(modulus);
            freebn(expected);
            freebn(answer);
        } else if (!strcmp(buf, "modsqrt")) {
            Bignum x, p, expected, answer;

            if (ptrnum != 3) {
                printf("%d: modsqrt with %d parameters, expected 3\n", line, ptrnum);
                exit(1);
            }

            x = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
            p = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
            expected = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
            answer = modsqrt(x, p);
            if (!answer)
                answer = copybn(Zero);

            if (bignum_cmp(expected, answer) == 0) {
                passes++;
            } else {
                char *xs = bignum_decimal(x);
                char *ps = bignum_decimal(p);
                char *qs = bignum_decimal(answer);
                char *ws = bignum_decimal(expected);

                printf("%d: fail: sqrt(%s) mod %s gave %s expected %s\n",
                       line, xs, ps, qs, ws);
                fails++;

                sfree(xs);
                sfree(ps);
                sfree(qs);
                sfree(ws);
            }
            freebn(p);
            freebn(x);
            freebn(expected);
            freebn(answer);
        } else {
            printf("%d: unrecognised test keyword: '%s'\n", line, buf);
            exit(1);
        }

        sfree(buf);
        sfree(data);
    }

    printf("passed %d failed %d total %d\n", passes, fails, passes+fails);
    return fails != 0;
}

#endif