/*
     * 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;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;

            while( max >= min ) {
                int mid = (min + max) / 2;

                if( errcode > NT_STATUS_CODES[mid] ) {
                    min = mid + 1;
                } else if( errcode < NT_STATUS_CODES[mid] ) {
                    max = mid - 1;
                } else {
                    return NT_STATUS_MESSAGES[mid];
                }
            }
        } else {
            int min = 0;

      while (low < high) {
        var mid = (low + high) / 2
        // Search for a '\n' that marks the start of a value. Don't go back past the start of the
        // array.
        while (mid > -1 && this[mid] != '\n'.code.toByte()) {
          mid--
        }
        mid++

        // Now look for the ending '\n'.
        var end = 1
        while (this[mid + end] != '\n'.code.toByte()) {
          end++

        while ( max >= min ) {
            int mid = ( min + max ) / 2;

            if ( errcode > DCERPC_FAULT_CODES[ mid ] ) {
                min = mid + 1;
            }
            else if ( errcode < DCERPC_FAULT_CODES[ mid ] ) {
                max = mid - 1;
            }
            else {
                return DCERPC_FAULT_MESSAGES[ mid ];
            }
        }

        this.flags &= ~flag;
    }


    /**
     * @return the mid
     */
    @Override
    public final long getMid () {
        return this.mid;
    }


    /**
     * @param mid
     *            the mid to set
     */
    @Override
    public final void setMid ( long mid ) {
        this.mid = mid;
    }


    /**
     * @param n

        while (max >= min) {
            int mid = (min + max) / 2;

            if (errcode > DCERPC_FAULT_CODES[mid]) {
                min = mid + 1;
            } else if (errcode < DCERPC_FAULT_CODES[mid]) {
                max = mid - 1;
            } else {
                return DCERPC_FAULT_MESSAGES[mid];
            }
        }

        return "0x" + Hexdump.toHexString(errcode, 8);

        return false;
    }


    /**
     * @return the mid
     */
    @Override
    public final long getMid () {
        return this.mid;
    }


    /**
     * @param mid
     *            the mid to set
     */
    @Override
    public final void setMid ( long mid ) {
        this.mid = (int) mid;
    }


    /**
     * @return the tid
     */

    /*
     * 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;
    for (int i = 0; i < reps; i++) {
      int j = i & ARRAY_MASK;

     */
    void setResponse ( CommonServerMessageBlockResponse msg );


    /**
     * 
     * @return the message id
     */
    long getMid ();


    /**
     * @param mid
     */
    void setMid ( long mid );


    /**
     * @return the command
     */
    int getCommand ();


    /**
     * @param command
     */
    void setCommand ( int command );


    /**

    boolean midLessThanFrom = (array[mid] < array[from]);
    boolean toLessThanFrom = (array[to] < array[from]);
    if (toLessThanMid == midLessThanFrom) {
      // Either array[to] < array[mid] < array[from] or array[from] <= array[mid] <= array[to].
      swap(array, mid, from);
    } else if (toLessThanMid != toLessThanFrom) {