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(unsigned long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  unsigned long long n, i;
  cin >> n;
  if (n & 1) {
    bool x = 1;
    for (i = 3; x && i * i <= n; i++) {
      if (n % i == 0) {
        if (isprime(i)) {
          x = 0;
          n = n - i;
        } else {
          if (isprime(n / i)) {
            x = 0;
            n = n - n / i;
          }
        }
      }
    }
    if (!x)
      cout << n / 2 + 1;
    else
      cout << "1";
  } else
    cout << n / 2;
}
