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() {
  unsigned long long int ans, num, i, even, odd, limit1, limit2;
  while (scanf("%llu", &num) == 1) {
    if (num % 2 == 0) {
      limit1 = num / 2;
      limit2 = limit1;
    } else if (num % 2 == 1) {
      limit1 = (num - 1) / 2;
      limit2 = limit1 + 1;
    }
    even = limit1 * (limit1 + 1);
    odd = limit2 * limit2;
    if (even > odd) {
      ans = even - odd;
      printf("%llu\n", ans);
    } else if (odd > even) {
      ans = odd - even;
      printf("-%llu\n", ans);
    } else
      printf("0\n");
  }
  return 0;
}
