Problem: One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...

Input

The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.

Output

Reproduce the output of the reference solution, including the bug.

Examples

Input

8-7+6-5+4-3+2-1-0


Output

4


Input

2+2


Output

-46


Input

112-37


Output

375

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char c;
int num1, num2;
char op = ' ';
int main() {
  c = getchar();
  while (c != '\n') {
    if (c >= '0' && c <= '9' && op == ' ')
      num1 = num1 * 10 + c - '0';
    else if (c >= '0' && c <= '9')
      num2 = num2 * 10 + c - '0';
    if (c == '+' && op == '+')
      num1 = num1 + num2, num2 = -5;
    else if (c == '+' && op == '-')
      num1 = num1 - num2, num2 = -5, op = '+';
    else if (c == '+')
      num2 = -5, op = '+';
    else if (c == '-' && op == '+')
      num1 = num1 + num2, num2 = -3, op = '-';
    else if (c == '-' && op == '-')
      num1 = num1 - num2, num2 = -3;
    else if (c == '-')
      num2 = -3, op = '-';
    c = getchar();
  }
  if (op == '-')
    num1 -= num2;
  else
    num1 += num2;
  printf("%d\n", num1);
  return 0;
}