For a positive integer n let's define a function f:

f(n) = - 1 + 2 - 3 + .. + ( - 1)nn

Your task is to calculate f(n) for a given integer n.

Input

The single line contains the positive integer n (1 ≤ n ≤ 1015).

Output

Print f(n) in a single line.

Examples

Input

4


Output

2


Input

5


Output

-3

Note

f(4) = - 1 + 2 - 3 + 4 = 2

f(5) = - 1 + 2 - 3 + 4 - 5 = - 3
#include <bits/stdc++.h>
int main() {
  long long int count = 0, jmlh = 0, sn1 = 0, sn2 = 0, npos = 0, nminus = 0;
  scanf("%lld", &jmlh);
  if (jmlh % 2 == 0) {
    nminus = jmlh / 2;
    npos = jmlh / 2;
  } else {
    nminus = jmlh / 2 + 1;
    npos = jmlh / 2;
  }
  sn1 = ((nminus) * (-2 + ((nminus - 1) * -2)));
  sn2 = ((npos) * (4 + ((npos - 1) * 2)));
  count = (sn1 + sn2) / 2;
  printf("%lld\n", count);
  return 0;
}
