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>
using namespace std;
long long n;
void output() {
  long long k = n / 2LL;
  if (n & 1LL) k = (k + 1) * -1;
  cout << k << "\n";
}
void input() { cin >> n; }
void solve() { return; }
void test_case() {
  input();
  output();
}
void pre() { return; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q, multicase = 0, precalc = 0;
  if (precalc) pre();
  if (multicase)
    cin >> q;
  else
    q = 1;
  for (int i = 1; i <= q; ++i) test_case();
}
