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

57785ca45b8873032f17 GCD = 42 A = 0 B = 42 LCM = 0 GCD = 42 A = 42 B = 0 LCM = 0 GCD = 42 A = 42 B = 42 LCM = 42 GCD = f60d A = ef7886c3391407529d5c B = d1d3ec32fa3103911830 LCM = cc376ed2dc362c38a45a GCD = 9370 A = 1ee02fb1c02100d1937f B = 67432fd1482d19c4a1c2 LCM = 159ff177bdb0ffbd09e2 GCD = c5f A = 5a3a2088b5c759420ed0 B = 1b1eb33b006a98178bb3 LCM = c5cbbbe9532d30d2a7dd GCD = e052 A = 67429f79b2ec3847cfc7 B = 39faa7cbdeb78f9028c1 LCM = 1ab071fb733ef142e94d GCD = 3523 A = 0 B = 3523 LCM = 0 GCD = 3523 A = 3523...