result += aHi * bLo; // aHi * bLo < 2^63, result < 2^64
        if (result < 0) {
          result = UnsignedLongs.remainder(result, m);
        }
        // result < 2^63 again
        result += aLo * bHi; // aLo * bHi < 2^63, result < 2^64
        result = times2ToThe32Mod(result, m); // result < m < 2^63

   */
  public static long divide(long dividend, long divisor) {
    if (divisor < 0) { // i.e., divisor >= 2^63:
      if (compare(dividend, divisor) < 0) {
        return 0; // dividend < divisor
      } else {
        return 1; // dividend >= divisor
      }
    }

    // Optimization - use signed division if dividend < 2^63
    if (dividend >= 0) {
      return dividend / divisor;
    }

    /*

    assertFalse(Utf8.isWellFormed(toByteArray(bytes)));
  }

  private static long[] generateFourByteShardsExpectedRunnables() {
    long[] expected = new long[128];
    // 0-63 are all 5300224
    for (int i = 0; i <= 63; i++) {
      expected[i] = 5300224;
    }
    // 97-111 are all 2342912
    for (int i = 97; i <= 111; i++) {
      expected[i] = 2342912;
    }
    // 113-117 are all 1048576

    // First add all the integer candidate values.
    longValues.addAll(Iterables.transform(POSITIVE_INTEGER_CANDIDATES, TO_LONG));
    // Add boundary values manually to avoid over/under flow (this covers 2^N for 31 and 63).
    longValues.add(Integer.MAX_VALUE + 1L, Long.MAX_VALUE - 1L, Long.MAX_VALUE);

    // Now add values near 2^N for lots of values of N.
    for (int exponent : asList(32, 33, 39, 40, 41, 47, 48, 49, 55, 56, 57)) {

    // parseInt blows up on overflow as opposed to intValue() which does not.
    return Long.parseLong(bigMean.toString());
  }

  private static boolean fitsInLong(BigInteger big) {
    return big.bitLength() <= 63;
  }

  private static final BigInteger MAX_LONG = BigInteger.valueOf(Long.MAX_VALUE);
  private static final BigInteger MIN_LONG = BigInteger.valueOf(Long.MIN_VALUE);

    // parseInt blows up on overflow as opposed to intValue() which does not.
    return Long.parseLong(bigMean.toString());
  }

  private static boolean fitsInLong(BigInteger big) {
    return big.bitLength() <= 63;
  }

  private static final BigInteger MAX_LONG = BigInteger.valueOf(Long.MAX_VALUE);
  private static final BigInteger MIN_LONG = BigInteger.valueOf(Long.MIN_VALUE);

          .add(BigInteger.valueOf(Long.MIN_VALUE).add(bigI))
          .add(BigInteger.valueOf(Integer.MAX_VALUE).add(bigI))
          .add(BigInteger.valueOf(Integer.MIN_VALUE).add(bigI))
          .add(BigInteger.ONE.shiftLeft(63).add(bigI))
          .add(BigInteger.ONE.shiftLeft(64).add(bigI));
    }
    TEST_LONGS = testLongsBuilder.build();
    TEST_BIG_INTEGERS = testBigIntegersBuilder.build();
  }

    assertFalse(Utf8.isWellFormed(toByteArray(bytes)));
  }

  private static long[] generateFourByteShardsExpectedRunnables() {
    long[] expected = new long[128];
    // 0-63 are all 5300224
    for (int i = 0; i <= 63; i++) {
      expected[i] = 5300224;
    }
    // 97-111 are all 2342912
    for (int i = 97; i <= 111; i++) {
      expected[i] = 2342912;
    }
    // 113-117 are all 1048576

    // First add all the integer candidate values.
    longValues.addAll(Iterables.transform(POSITIVE_INTEGER_CANDIDATES, TO_LONG));
    // Add boundary values manually to avoid over/under flow (this covers 2^N for 31 and 63).
    longValues.add(Integer.MAX_VALUE + 1L, Long.MAX_VALUE - 1L, Long.MAX_VALUE);

    // Now add values near 2^N for lots of values of N.
    for (int exponent : asList(32, 33, 39, 40, 41, 47, 48, 49, 55, 56, 57)) {

          .add(BigInteger.valueOf(Long.MIN_VALUE).add(bigI))
          .add(BigInteger.valueOf(Integer.MAX_VALUE).add(bigI))
          .add(BigInteger.valueOf(Integer.MIN_VALUE).add(bigI))
          .add(BigInteger.ONE.shiftLeft(63).add(bigI))
          .add(BigInteger.ONE.shiftLeft(64).add(bigI));
    }
    TEST_LONGS = testLongsBuilder.build();
    TEST_BIG_INTEGERS = testBigIntegersBuilder.build();
  }