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() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    long long n, c = 0;
    cin >> n;
    while (1) {
      if (n == 0) break;
      int i, p = 0;
      for (i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) {
          p = 1;
          break;
        }
      }
      if (p == 1)
        n = n - n / i;
      else
        n = 0;
      c++;
    }
    cout << c << endl;
  }
  return 0;
}
