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;
const int lim = 1e5 + 5;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  if (n % 2 != 0)
    for (int i = 2; i < lim; ++i)
      if (n % i == 0) {
        n = n - i + 2;
        break;
      }
  if (n % 2 != 0)
    cout << 1;
  else
    cout << n / 2;
  return 0;
}
