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>
int main() {
  int num;
  long long n;
  scanf("%lld", &n);
  int i = 2;
  while ((n % i) != 0 && i * 1.0 < sqrt(n)) i++;
  if (i * 1.0 > sqrt(n)) {
    num = 1;
  } else {
    num = (n - i) / 2 + 1;
  }
  printf("%d", num);
}
