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;
long long inf = 1e9 + 7;
const long long maxinput = 100005;
int main() {
  long long n;
  cin >> n;
  for (long long i = 2; i < maxinput; i++) {
    if ((n % i) == 0LL) {
      if (i == 2)
        cout << n / i << "\n";
      else
        cout << 1 + ((n - i) / 2) << "\n";
      return (0);
    }
  }
  cout << 1 << "\n";
}
