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 isprime(long long int x) {
  for (int p = 2; p <= sqrt(x); p++) {
    if (x % p == 0) return 0;
  }
  return 1;
}
int main() {
  long long int n, t, k;
  while (~scanf("%lld", &n) && n) {
    t = 0;
    if (isprime(n)) {
      printf("1\n");
      continue;
    }
    if (n % 2 != 0) {
      for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) {
          if (isprime(i)) {
            k = i;
            break;
          }
        }
      }
      n = n - k;
      t++;
    }
    t += n / 2;
    printf("%lld\n", t);
  }
  return 0;
}
