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

      for (int b : POSITIVE_INTEGER_CANDIDATES) {
        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(IntMath.gcd(a, b)));
      }
    }
  }

  public void testGCDZero() {
    for (int a : POSITIVE_INTEGER_CANDIDATES) {
      assertEquals(a, IntMath.gcd(a, 0));
      assertEquals(a, IntMath.gcd(0, a));
    }
    assertEquals(0, IntMath.gcd(0, 0));
  }

  public void testGCDNegativePositiveThrows() {

        return DoubleMath.factorial(n);
      }

      @Override
      public int gcdInt(int a, int b) {
        return IntMath.gcd(a, b);
      }

      @Override
      public long gcdLong(long a, long b) {
        return LongMath.gcd(a, b);
      }

      @Override
      public long binomialCoefficient(int n, int k) {
        return LongMath.binomial(n, k);
      }

      @Override

        assertEquals(valueOf(a).gcd(valueOf(b)), valueOf(LongMath.gcd(a, b)));
      }
    }
  }

  @GwtIncompatible // TODO
  public void testGCDZero() {
    for (long a : POSITIVE_LONG_CANDIDATES) {
      assertEquals(a, LongMath.gcd(a, 0));
      assertEquals(a, LongMath.gcd(0, a));
    }
    assertEquals(0, LongMath.gcd(0, 0));
  }

  @GwtIncompatible // TODO

  int mod(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += LongMath.mod(longs[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark
  int gCD(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += LongMath.mod(nonnegative[j], positive[j]);
    }
    return tmp;
  }

  @Benchmark

    //     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