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 long long N = 100005;
void inp_out() {}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  inp_out();
  {
    long long i, j, k, n, m, ans = 1e18, cnt = 0, sum = 0;
    cin >> n;
    if (n % 2 == 0) {
      cout << n / 2 << "\n";
      return 0;
    }
    for (i = 1; i * i <= n; i++) {
      if (n % i == 0) {
        if (i != 1) {
          ans = min(ans, i);
        }
        ans = min(ans, n / i);
      }
    }
    n -= ans;
    cnt++;
    cout << cnt + n / 2 << "\n";
  }
  return 0;
}
