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;
  cin >> n;
  long long c = 0;
  while (n) {
    if (n % 2 == 0) {
      c += n / 2;
      n = 0;
    }
    for (int i = 3; i <= sqrt(n); i += 2) {
      while (n % i == 0 && n) {
        c += n / i;
        n = 0;
      }
      if (n == 0) break;
    }
    if (n > 2) {
      n -= n;
      c++;
    }
  }
  cout << c;
  return 0;
}
