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;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int first_prime(long long int num) {
  for (long long int i = 2; i < sqrt(num); i++) {
    if (num % i == 0) {
      if (isPrime(i))
        ;
      return (i);
    }
  }
}
int main() {
  long long int n, ans = 0;
  cin >> n;
  if (n % 2 == 0)
    ans = n / 2;
  else if (isPrime(n))
    ans = 1;
  else {
    ans++;
    long long int tmp = first_prime(n);
    n = n - tmp;
    ans += n / 2;
  }
  cout << ans << endl;
  return 0;
}
