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;
int i, j, k;
int main() {
  unsigned long long n;
  scanf("%lld", &n);
  unsigned long long cnt = 0;
  if (n % 2 == 0)
    cnt = n / 2;
  else {
    unsigned long long lmt = sqrt(1 + n);
    for (i = 3; i <= lmt; i++) {
      if (n % i == 0) {
        n -= i;
        cnt = n / 2 + 1;
        break;
      }
    }
    if (cnt == 0) cnt = 1;
  }
  printf("%lld\n", cnt);
}
