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() {
  long long n, i, l, a = 0, p;
  scanf("%lld", &n);
  while (n % 2 != 0) {
    p = n;
    for (i = 2; i * i < n; i++)
      if (n % i == 0) {
        p = i;
        break;
      }
    n -= p;
    a++;
  }
  a += n / 2;
  printf("%lld\n", a);
  return 0;
}
