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 i, n, d, j, dem = 0, k;
int main() {
  scanf("%lld\n", &n);
  long long tam = n;
  i = 1;
  while (n > 0) {
    i++;
    if (i > sqrt(tam) && n == tam) {
      dem = 1;
      break;
    }
    if (n % i == 0) {
      dem = 1 + (n - i) / 2;
      break;
    }
  }
  cout << dem;
}
