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 < 2) return 0;
  long long int i = 2;
  while (i * i <= n) {
    if (n % i == 0) return 0;
    i++;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
  } else {
    if (isprime(n)) {
      cout << 1;
      return 0;
    }
    long long int i = 3;
    for (; i * i < n; i += 2) {
      if (n % i == 0) {
        break;
      }
    }
    cout << (n - i) / 2 + 1;
  }
  return 0;
}
