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() {
  unsigned long long n;
  cin >> n;
  unsigned long long count = 0;
  while (n > 0 && n % 2) {
    unsigned long long d;
    bool prime = true;
    for (d = 3; d * d <= n && n % d; d += 2) prime = false;
    if (prime) d = n;
    n -= d;
    count++;
  }
  count += n / 2;
  cout << count << '\n';
}
