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;
template <typename T>
T gcd(T a, T b) {
  return a ? gcd(b % a, a) : b;
}
template <typename T>
T lcm(T a, T b) {
  return (a / gcd(a, b)) * b;
}
bool is_down(char x) { return ('a' <= x && x <= 'z'); }
bool is_upper(char x) { return ('A' <= x && x <= 'Z'); }
bool is_digit(char x) { return ('0' <= x && x <= '9'); }
const long double pi = 3.141592653589793238462643383279;
const long double log23 = 1.58496250072115618145373894394781;
const long double eps = 1e-10;
const long double zero = 0;
const long long INF = 1e18;
const int COUT = 30;
const int prost = 239;
const long long prost64 = 239;
const int two = 2;
const int thr = 3;
const long long sr = 31;
const long long MOD = 1e9 + 7;
const long long MOD2 = MOD * MOD;
const int BIG = 1e9 + 239;
const int alf = 26;
const int M = 2 * 1e5 + 239;
const int N = 2 * 1e3 + 239;
const int T = (1 << 20);
const int B = trunc(sqrt(M)) + 1;
const int MAX_LOG = 19;
const int km = (1 << 18);
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int bit_max = 32;
const int dig = 10;
const string str_alf = "abcdefghijklmnopqrstuvwxyz";
const string str_alf_big = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const int day[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int digarr[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
const int bt = 31;
int main() {
  cin.sync_with_stdio(0);
  string s;
  cin >> s;
  int ans = 0;
  int now = 0;
  int sg = 1;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == '-' || s[i] == '+') {
      ans += sg * now;
      now = 0;
    }
    if (s[i] == '-')
      sg = -1;
    else if (s[i] == '+')
      sg = 1;
    now = 10 * now + (s[i] - '0');
  }
  ans += sg * now;
  cout << ans;
  return 0;
}