*
   * <p>This method is suitable for the common use case of dividing work among buckets that meet the
   * following conditions:
   *
   * <ul>
   *   <li>You want to assign the same fraction of inputs to each bucket.
   *   <li>When you reduce the number of buckets, you can accept that the most recently added
   *       buckets will be removed first. More concretely, if you are dividing traffic among tasks,

   *     partitions
   * @throws IllegalArgumentException if {@code size} is nonpositive
   */
  public static <T extends @Nullable Object> UnmodifiableIterator<List<T>> partition(
      Iterator<T> iterator, int size) {
    return partitionImpl(iterator, size, false);
  }

  /**

        for (RoundingMode mode : ALL_SAFE_ROUNDING_MODES) {
          BigInteger expected =
              new BigDecimal(p).divide(new BigDecimal(q), 0, mode).toBigIntegerExact();
          assertEquals(expected, BigIntegerMath.divide(p, q, mode));
        }
      }
    }
  }

  private static final BigInteger BAD_FOR_ANDROID_P = new BigInteger("-9223372036854775808");

  }

  /**
   * Verifies that the specified expected result is always produced by collecting the specified
   * inputs, regardless of how the elements are divided.
   */
  @SafeVarargs
  @CanIgnoreReturnValue
  @SuppressWarnings("nullness") // TODO(cpovirk): Remove after we fix whatever the bug is.

      int j = i & ARRAY_MASK;
      tmp += IntMath.sqrt(positive[j], mode);
    }
    return tmp;
  }

  @Benchmark
  int divide(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += IntMath.divide(ints[j], nonzero[j], mode);
    }
    return tmp;
  }

      int j = i & ARRAY_MASK;
      tmp += BigIntegerMath.sqrt(positive[j], mode).intValue();
    }
    return tmp;
  }

  @Benchmark
  int divide(int reps) {
    int tmp = 0;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;
      tmp += BigIntegerMath.divide(nonzero1[j], nonzero2[j], mode).intValue();
    }
    return tmp;
  }

  @Benchmark
  long roundToDouble(int reps) {
    long tmp = 0;

    new RoundToDoubleTester(BigDecimal.ZERO).setExpectation(0.0, RoundingMode.values()).test();
  }

  public void testRoundToDouble_oneThird() {
    new RoundToDoubleTester(
            BigDecimal.ONE.divide(BigDecimal.valueOf(3), new MathContext(50, HALF_EVEN)))
        .roundUnnecessaryShouldThrow()
        .setExpectation(0.33333333333333337, UP, CEILING)

          }
          int expected =
              new BigDecimal(valueOf(p)).divide(new BigDecimal(valueOf(q)), 0, mode).intValue();
          assertEquals(p + "/" + q, force32(expected), IntMath.divide(p, q, mode));
          // Check the assertions we make in the javadoc.
          if (mode == DOWN) {
            assertEquals(p + "/" + q, p / q, IntMath.divide(p, q, mode));
          } else if (mode == FLOOR) {

  }

  /**
   * Binary search [sections] for [codePoint], looking at its top 14 bits.
   *
   * This binary searches over 4-byte entries, and so it needs to adjust binary search indices
   * in (by dividing by 4) and out (by multiplying by 4).
   */
  private fun findSectionsIndex(codePoint: Int): Int {
    val target = (codePoint and 0x1fff80) shr 7
    val offset =
      binarySearch(
        position = 0,

        result *= i;
      }
      return BigInteger.valueOf(result);
    }

    /*
     * We want each multiplication to have both sides with approximately the same number of digits.
     * Currently, we just divide the range in half.
     */
    int mid = (n1 + n2) >>> 1;
    return oldSlowFactorial(n1, mid).multiply(oldSlowFactorial(mid, n2));
  }

  @Benchmark
  int slowFactorial(int reps) {
    int tmp = 0;