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 n) {
  double n0 = (double)n;
  long long sqn = sqrt(n0);
  sqn += 2;
  for (long long i = 2; i <= sqn; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long divisor(long long n) {
  for (long long i = 2; i < n; i++) {
    if (n % i == 0) {
      return i;
    }
  }
}
int main() {
  long long n;
  while (scanf("%I64d", &n) != EOF) {
    if (n % 2 == 0)
      printf("%I64d\n", n / 2);
    else {
      if (isprime(n)) {
        printf("1\n");
      } else {
        n -= divisor(n);
        printf("%I64d\n", n / 2 + 1);
      }
    }
  }
  return 0;
}
