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 i = 2;
  long long lim = (long long)sqrt(n + 0.0);
  bool f = false;
  long long ans = 0;
  while (true) {
    long long t = -1;
    if (n % i == 0) {
      long long r = n / i;
      ans += r - i + 1;
      n = (i - 1) * i;
      t = (i - 2);
      i = 1;
      f = true;
    }
    if (t == 0) {
      ans++;
      break;
    }
    i++;
  }
  if (!f) ans = 1;
  cout << ans;
  return 0;
}
