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 mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  if (n % 2 == 0)
    cout << n / 2;
  else {
    bool found = false;
    long long ans = 0;
    for (int i = 3; i <= 1e5 + 30; i += 2) {
      if (n % i == 0) {
        n -= i;
        found = true;
        break;
      }
    }
    if (!found)
      ans = 1;
    else
      ans = 1 + (n) / 2;
    cout << ans;
  }
}
