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 is_p(long long int a) {
  int sq = sqrt(a);
  for (int i = 2; i <= sq; i++)
    if (a % i == 0) return i;
  return -1;
}
int main() {
  long long int n, ans, x;
  while (cin >> x) {
    if (1 & x) {
      n = is_p(x);
      if (n == -1)
        cout << "1" << endl;
      else {
        x -= n;
        ans = x / 2;
        cout << ans + 1 << endl;
      }
    } else
      cout << x / 2 << endl;
  }
}
