assertEquals(0, Integer.numberOfTrailingZeros(ACE.FILE_READ_DATA), "FILE_READ_DATA should be at bit position 0");
            assertEquals(1, Integer.numberOfTrailingZeros(ACE.FILE_WRITE_DATA), "FILE_WRITE_DATA should be at bit position 1");
            assertEquals(28, Integer.numberOfTrailingZeros(ACE.GENERIC_ALL), "GENERIC_ALL should be at bit position 28");

    int startingNumber = LongMath.factorials.length;
    long product = LongMath.factorials[startingNumber - 1];
    // Strip off 2s from this value.
    int shift = Long.numberOfTrailingZeros(product);
    product >>= shift;

    // Use floor(log2(num)) + 1 to prevent overflow of multiplication.
    int productBits = LongMath.log2(product, FLOOR) + 1;

     * >60% faster than the Euclidean algorithm in benchmarks.
     */
    int aTwos = Long.numberOfTrailingZeros(a);
    a >>= aTwos; // divide out all 2s
    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:

    } else {
      char[] buf = new char[64];
      int i = buf.length;
      if ((radix & (radix - 1)) == 0) {
        // Radix is a power of two so we can avoid division.
        int shift = Integer.numberOfTrailingZeros(radix);
        int mask = radix - 1;
        do {
          buf[--i] = Character.forDigit(((int) x) & mask, radix);
          x >>>= shift;
        } while (x != 0);
      } else {

        @Override
        public boolean hasNext() {
          return remainingSetBits != 0;
        }

        @Override
        public E next() {
          int index = Integer.numberOfTrailingZeros(remainingSetBits);
          if (index == 32) {
            throw new NoSuchElementException();
          }
          remainingSetBits &= ~(1 << index);
          return elements.get(index);
        }