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 main() {
  long long int n, i, j;
  cin >> n;
  vector<int> k, a;
  for (i = 0; i <= 100000; i++) a.push_back(0);
  a[0] = 1;
  a[1] = 1;
  for (i = 2; i < 100000; i++) {
    if (a[i] == 0) {
      for (j = i; j < 100000; j = j + i) {
        a[j] = 1;
      }
      k.push_back(i);
    }
  }
  for (i = 0; i < k.size(); i++) {
    if (n % k[i] == 0) {
      if (k[i] == 2) {
        cout << n / 2 << endl;
        return 0;
      }
      if (k[i] != 2) {
        n -= k[i];
        cout << 1 + n / 2 << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl;
  return 0;
}
