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;
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
}
void solve() {
  long long n;
  cin >> n;
  long long res = 0;
  while (n != 0) {
    if (n % 2 == 0) {
      res += n / 2;
      n = 0;
      break;
    }
    bool found = false;
    for (long long i = 2; i * i <= n; ++i) {
      if (n % i == 0) {
        n -= i;
        ++res;
        found = true;
        break;
      }
    }
    if (!found) {
      ++res;
      break;
    }
  }
  cout << res << '\n';
  ;
}
