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(void) {
  long long int n, i;
  scanf("%lld", &n);
  long long int count = 0;
  if (!(n & 1))
    count = n / 2;
  else {
    for (i = 3; i * i <= n; i += 2) {
      if (!(n % i)) {
        count++;
        n -= i;
        break;
      }
    }
    if (n & 1) {
      n = 0;
      count++;
    }
    count += n / 2;
  }
  printf("%lld\n", count);
  return 0;
}
