"type":       "stop",

          A result(Collector<T, A, R> collector, Iterable<T> inputs) {
        A accum = collector.supplier().get();
        for (T input : inputs) {
          A newAccum = collector.supplier().get();
          collector.accumulator().accept(newAccum, input);
          accum = collector.combiner().apply(accum, newAccum);
        }
        return accum;
      }
    },

          A result(Collector<T, A, R> collector, Iterable<T> inputs) {
        A accum = collector.supplier().get();
        for (T input : inputs) {
          A newAccum = collector.supplier().get();
          collector.accumulator().accept(newAccum, input);
          accum = collector.combiner().apply(accum, newAccum);
        }
        return accum;
      }
    },

                }),
        (accum, t) -> {
          /*
           * We assign these to variables before calling checkNotNull to work around a bug in our
           * nullness checker.
           */
          K key = keyFunction.apply(t);
          V newValue = valueFunction.apply(t);
          accum.put(
              checkNotNull(key, "Null key for input %s", t),

                }),
        (accum, t) -> {
          /*
           * We assign these to variables before calling checkNotNull to work around a bug in our
           * nullness checker.
           */
          K key = keyFunction.apply(t);
          V newValue = valueFunction.apply(t);
          accum.put(
              checkNotNull(key, "Null key for input %s", t),

        default:
          throw new AssertionError();
      }
    }
    long accum = 1;
    while (true) {
      switch (k) {
        case 0:
          return accum;
        case 1:
          return checkedMultiply(accum, b);
        default:
          if ((k & 1) != 0) {
            accum = checkedMultiply(accum, b);
          }
          k >>= 1;
          if (k > 0) {
            checkNoOverflow(

      if (numeratorBits + bits >= Long.SIZE - 1) {
        // The numerator is as big as it can get without risking overflow.
        // Multiply numeratorAccum / denominatorAccum into accum.
        accum =
            accum
                .multiply(BigInteger.valueOf(numeratorAccum))
                .divide(BigInteger.valueOf(denominatorAccum));
        numeratorAccum = p;
        denominatorAccum = q;

    if (digit < 0 || digit >= radix) {
      return null;
    }
    long accum = -digit;

    long cap = Long.MIN_VALUE / radix;

    while (index < string.length()) {
      digit = AsciiDigits.digit(string.charAt(index++));
      if (digit < 0 || digit >= radix || accum < cap) {
        return null;
      }
      accum *= radix;
      if (accum < Long.MIN_VALUE + digit) {
        return null;
      }

  @GwtIncompatible // TODO
  private long simpleBinomial(int n, int k) {
    long accum = 1;
    for (int i = 0; i < k; i++) {
      accum = LongMath.checkedMultiply(accum, n - i);
      accum /= i + 1;
    }
    return accum;
  }

  @GwtIncompatible // java.math.BigInteger
  public void testIsPowerOfTwo() {
    for (long x : ALL_LONG_CANDIDATES) {

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