* >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:

    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 {

      // The logic here would be wrong for bitsPerChar > 8, but since we require distinct ASCII
      // characters that can't happen.
      int zeroesInBitsPerChar = Integer.numberOfTrailingZeros(bitsPerChar);
      this.charsPerChunk = 1 << (3 - zeroesInBitsPerChar);
      this.bytesPerChunk = bitsPerChar >> zeroesInBitsPerChar;

      this.mask = chars.length - 1;

      this.decodabet = decodabet;