.add(i)
          .add(Long.MAX_VALUE + i)
          .add(Long.MIN_VALUE + i)
          .add(Integer.MIN_VALUE + i)
          .add(Integer.MAX_VALUE + i)
          .add(floatConversionTest + i)
          .add(doubleConversionTest + i);
      BigInteger bigI = BigInteger.valueOf(i);
      testBigIntegersBuilder
          .add(bigI)
          .add(BigInteger.valueOf(Long.MAX_VALUE).add(bigI))

        return Long.MAX_VALUE;
      }
      if (end < start && result > 0) { // underflow
        return Long.MIN_VALUE;
      }
      return result;
    }

    @Override
    public Long minValue() {
      return Long.MIN_VALUE;
    }

    @Override
    public Long maxValue() {
      return Long.MAX_VALUE;
    }

    private Object readResolve() {

  /**
   * Returns the number of nanoseconds of the given duration without throwing or overflowing.
   *
   * <p>Instead of throwing {@link ArithmeticException}, this method silently saturates to either
   * {@link Long#MAX_VALUE} or {@link Long#MIN_VALUE}. This behavior can be useful when decomposing
   * a duration in order to call a legacy API which requires a {@code long, TimeUnit} pair.
   */
  static long toNanosSaturated(Duration duration) {

        SMBUtil.writeInt4(Integer.MAX_VALUE, buffer, bufferIndex + 3);
        SMBUtil.writeInt4(Integer.MAX_VALUE, buffer, bufferIndex + 7);
        SMBUtil.writeInt4(Integer.MAX_VALUE, buffer, bufferIndex + 11);
        SMBUtil.writeInt4(Integer.MAX_VALUE, buffer, bufferIndex + 15);
        SMBUtil.writeInt4(Integer.MAX_VALUE, buffer, bufferIndex + 19);
        SMBUtil.writeInt4(Integer.MAX_VALUE, buffer, bufferIndex + 23);

      @Nullable C lastValue;

      @Override
      protected @Nullable C computeNext() {
        while (merged.hasNext()) {
          C next = merged.next();
          boolean duplicate = lastValue != null && comparator.compare(next, lastValue) == 0;

          // Keep looping till we find a non-duplicate value.
          if (!duplicate) {
            lastValue = next;
            return lastValue;

            byte[] buffer = createValidWriteResponse(Integer.MAX_VALUE, Integer.MAX_VALUE);

            int bytesRead = response.readBytesWireFormat(buffer, 0);

            assertEquals(16, bytesRead);
            assertEquals(Integer.MAX_VALUE, response.getCount());
            assertEquals(Integer.MAX_VALUE, response.getRemaining());
        }

        @Test

    Double.NEGATIVE_INFINITY,
    -Double.MAX_VALUE,
    (double) Long.MIN_VALUE,
    (double) Integer.MIN_VALUE,
    -Math.PI,
    -1.0,
    -Double.MIN_VALUE,
    -0.0,
    +0.0,
    Double.MIN_VALUE,
    1.0,
    Math.PI,
    (double) Integer.MAX_VALUE,
    (double) Long.MAX_VALUE,
    Double.MAX_VALUE,
    Double.POSITIVE_INFINITY,
    Double.NaN,
    Float.MAX_VALUE,
  };

      @Nullable C lastValue;

      @Override
      protected @Nullable C computeNext() {
        while (merged.hasNext()) {
          C next = merged.next();
          boolean duplicate = lastValue != null && comparator.compare(next, lastValue) == 0;

          // Keep looping till we find a non-duplicate value.
          if (!duplicate) {
            lastValue = next;
            return lastValue;

        problem = new DefaultProblem(null, null, null, 42, -1, null);
        assertEquals(42, problem.getLineNumber());

        problem = new DefaultProblem(null, null, null, Integer.MAX_VALUE, -1, null);
        assertEquals(Integer.MAX_VALUE, problem.getLineNumber());

        // this case is not specified, might also return -1
        problem = new DefaultProblem(null, null, null, Integer.MIN_VALUE, -1, null);

   * Duplicated from LongMath.
   * binomial(biggestBinomials[k], k) fits in a long, but not
   * binomial(biggestBinomials[k] + 1, k).
   */
  static final int[] biggestBinomials = {
    Integer.MAX_VALUE,
    Integer.MAX_VALUE,
    Integer.MAX_VALUE,
    3810779,
    121977,
    16175,
    4337,
    1733,
    887,
    534,
    361,
    265,
    206,
    169,
    143,
    125,
    111,
    101,