int bTwos = Integer.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

    int bTwos = Long.numberOfTrailingZeros(b);
    b >>= bTwos; // divide out all 2s
    while (a != b) { // both a, b are odd
      // The key to the binary GCD algorithm is as follows:
      // Both a and b are odd. Assume a > b; then gcd(a - b, b) = gcd(a, b).
      // But in gcd(a - b, b), a - b is even and b is odd, so we can divide out powers of two.

      // We bend over backwards to avoid branching, adapting a technique from

    //     moved at that point. Otherwise, we can rotate the cycle a[1], a[1 + d], a[1 + 2d], etc,
    //     then a[2] etc, and so on until we have rotated all elements. There are gcd(d, n) cycles
    //     in all.
    // (3) "Successive". We can consider that we are exchanging a block of size d (a[0..d-1]) with a
    //     block of size n-d (a[d..n-1]), where in general these blocks have different sizes. If we