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