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;
int snt(long long soA) {
  if (soA < 2) return 0;
  for (long long i = 2; i <= sqrt((float)soA); i++) {
    if (soA % i == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  long long n, i, min;
  cin >> n;
  if (snt(n)) {
    cout << 1;
    return 0;
  }
  for (long long i = 2; i <= n; i++) {
    if (snt(i) && n % i == 0) {
      cout << n / i;
      return 0;
    }
  }
  if (n == 0) cout << 0;
}
