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>
int main() {
  long long n;
  while (~scanf("%lld", &n)) {
    long long is_odd = 0;
    if (n % 2 == 1) {
      is_odd = 1;
      bool flag = false;
      for (int i = 2; 1LL * i * i <= n && !flag; ++i) {
        if (n % i == 0) {
          n -= i;
          flag = true;
        }
      }
      if (!flag) n = 0;
    }
    printf("%lld\n", is_odd + n / 2);
  }
  return 0;
}
