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;
long long int a[100005] = {};
long long int find(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return n;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int i, j, k, n, m, ans = 0, cnt = 0, sum = 0;
    cin >> n;
    ans = find(n);
    if (ans == 2) {
      cout << n / ans;
    } else {
      n -= ans;
      cout << n / 2 + 1;
    }
  }
}
