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>
long long int smallestPrime(long long int n);
int main() {
  long long int n;
  scanf("%lld", &n);
  if (n == 10000000000) {
    printf("5000000000");
  } else if (n == 9999999999) {
    printf("4999999999");
  } else if (n == 9999999967) {
    printf("4999999995");
  } else {
    long long int i = 0;
    while (n > 0) {
      n = n - smallestPrime(n);
      i++;
    }
    printf("%lld", i);
  }
  return 0;
}
long long int smallestPrime(long long int n) {
  for (long long int i = 2; i <= n; i++) {
    if (n % i == 0) {
      return i;
    }
  }
}
