return entry.value;
    }
    entry = cacheEntry2;
    if (entry != null && entry.key == key) {
      // Promote second cache entry to first so the access pattern
      // [K1, K2, K1, K3, K1, K4...] still hits the cache half the time.
      addToCache(entry);
      return entry.value;
    }
    return null;
  }

  @Override
  void clearCache() {
    super.clearCache();
    cacheEntry1 = null;
    cacheEntry2 = null;
  }

FE20..FE23    ; valid                                  # 1.1  COMBINING LIGATURE LEFT HALF..COMBINING DOUBLE TILDE RIGHT HALF
FE24..FE26    ; valid                                  # 5.1  COMBINING MACRON LEFT HALF..COMBINING CONJOINING MACRON
FE27..FE2D    ; valid                                  # 7.0  COMBINING LIGATURE LEFT HALF BELOW..COMBINING CONJOINING MACRON BELOW

  }

  private static long random() {
    return randomSource.nextLong();
  }

  // A random value that cannot be 0 and that is unsigned-less-than or equal
  // to the given dividend, so that we don't have half of our divisions being
  // trivial because the divisor is bigger than the dividend.
  // Using remainder here does not give us a uniform distribution but it should
  // not have a big impact on the measurement.

            return roundFloorAsDouble;
          } else if (diff > 0) { // closer to ceiling
            return roundCeilingAsDouble;
          }
          // halfway between the representable values; do the half-whatever logic
          switch (mode) {
            case HALF_EVEN:
              // roundFloorAsDouble and roundCeilingAsDouble are neighbors, so precisely
              // one of them should have an even long representation

      }
      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;

  static <R, C, V> RegularImmutableTable<R, C, V> forOrderedComponents(
      ImmutableList<Cell<R, C, V>> cellList,
      ImmutableSet<R> rowSpace,
      ImmutableSet<C> columnSpace) {
    // use a dense table if more than half of the cells have values
    // TODO(gak): tune this condition based on empirical evidence
    return (cellList.size() > (((long) rowSpace.size() * columnSpace.size()) / 2))

   * gigantic heap, in which case we scale by heap size.
   */
  private static long timeoutSeconds() {
    // This class can make no hard guarantees.  The methods in this class are inherently flaky, but
    // we try hard to make them robust in practice.  We could additionally try to add in a system
    // load timeout multiplier.  Or we could try to use a CPU time bound instead of wall clock time

  }

  private static long random() {
    return randomSource.nextLong();
  }

  // A random value that cannot be 0 and that is unsigned-less-than or equal
  // to the given dividend, so that we don't have half of our divisions being
  // trivial because the divisor is bigger than the dividend.
  // Using remainder here does not give us a uniform distribution but it should
  // not have a big impact on the measurement.

"berori", "beroriek", "beste", "bezala", "da", "dago", "dira", "ditu", "du", "dute", "edo", "egin", "ere", "eta", "eurak", "ez", "gainera", "gu", "gutxi", "guzti", "haiei", "haiek", "haietan", "hainbeste", "hala", "han", "handik", "hango", "hara", "hari", "hark", "hartan", "hau", "hauei", "hauek", "hauetan", "hemen", "hemendik", "hemengo", "hi", "hona", "honek", "honela", "honetan", "honi", "hor", "hori", "horiei", "horiek", "horietan", "horko", "horra", "horrek", "horrela", "horretan", "horri", "hortik",...

"berori", "beroriek", "beste", "bezala", "da", "dago", "dira", "ditu", "du", "dute", "edo", "egin", "ere", "eta", "eurak", "ez", "gainera", "gu", "gutxi", "guzti", "haiei", "haiek", "haietan", "hainbeste", "hala", "han", "handik", "hango", "hara", "hari", "hark", "hartan", "hau", "hauei", "hauek", "hauetan", "hemen", "hemendik", "hemengo", "hi", "hona", "honek", "honela", "honetan", "honi", "hor", "hori", "horiei", "horiek", "horietan", "horko", "horra", "horrek", "horrela", "horretan", "horri", "hortik",...