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 main() {
  long long n;
  scanf("%lld", &n);
  if (n % 2 == 0)
    printf("%lld", (n / 2));
  else {
    vector<bool> nums(n + 1, true);
    for (long long p = 2; p * p <= n; p++) {
      if (nums[p]) {
        if (n % p == 0) {
          n -= p;
          break;
        } else {
          for (long long i = p * 2; i <= n; i += p) nums[i] = false;
        }
      }
    }
    printf("%lld", 1 + (n / 2));
  }
  return 0;
}
