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 is_prime(long long n) {
  if (n % 2 == 0) return false;
  for (long long i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
int main() {
  long long n;
  cin >> n;
  if (is_prime(n)) {
    cout << 1 << endl;
    return 0;
  } else if (n % 2 == 0) {
    cout << n / 2 << endl;
    return 0;
  } else {
    long long c = 0, i;
    while (n >= 0) {
      for (i = 2; i * i <= n; ++i) {
        if (is_prime(i)) {
          if (n % i == 0) break;
        }
      }
      n -= i;
      c++;
      if (n % 2 == 0) {
        cout << c + n / 2 << endl;
        return 0;
      }
    }
    cout << c << endl;
    return 0;
  }
}
