// u v = SUM_{0<=i,j<n} u_i v_j t^{i+j}
//
// Suppose instead that we're given ũ = SUM_{0<=i<n} u_{n-i-1} t^i
- // and ṽ = SUM_{0<=j<n} v_{n-j-1} t^j, so the bits are backwards.
+ // and ṽ = SUM_{0<=j<n} v_{n-j-1} t^j, so the bits are backwards.
// Then
//
- // ũ ṽ = SUM_{0<=i,j<n} u_{n-i-1} v_{n-j-1} t^{i+j}
+ // ũ ṽ = SUM_{0<=i,j<n} u_{n-i-1} v_{n-j-1} t^{i+j}
// = SUM_{0<=i,j<n} u_i v_j t^{2n-2-(i+j)}
//
// which is almost the bit-reversal of u v, only it's shifted right
//
// q = r s = (u_0 + u_1) (v_0 + v_1)
// = (u_0 v_0) + (u1 v_1) + (u_0 v_1 + u_1 v_0)
- // = a + d + c
+ // = a + c + b
//
// The first two terms we've already calculated; the last is the
// remaining one we want. We'll set B = t^128. We know how to do