You are given an integer number n. The following algorithm is applied to it:

  1. if n = 0, then end algorithm; 
  2. find the smallest prime divisor d of n; 
  3. subtract d from n and go to step 1. 



Determine the number of subtrations the algorithm will make.

Input

The only line contains a single integer n (2 ≤ n ≤ 10^{10}).

Output

Print a single integer — the number of subtractions the algorithm will make.

Examples

Input


5


Output


1


Input


4


Output


2

Note

In the first example 5 is the smallest prime divisor, thus it gets subtracted right away to make a 0.

In the second example 2 is the smallest prime divisor at both steps.
#include <bits/stdc++.h>
using namespace std;
long long H, cur, cnt;
int main() {
  cin >> H;
  while (H) {
    bool ok = false;
    if (H % 2 == 0) {
      cnt += H / 2;
      cnt--;
      break;
    }
    for (int i = 2; i <= sqrt(H); i++) {
      if (H % i == 0) {
        ok = true;
        cur = i;
        break;
      }
    }
    if (!ok) break;
    H -= cur;
    cnt++;
  }
  cout << cnt + 1 << endl;
  return 0;
}
