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 n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  }
  int sn = (int)sqrt(n) + 1;
  bool good = false;
  for (int i = 2; i < sn; ++i) {
    if (n % i == 0) {
      n -= i;
      good = true;
      break;
    }
  }
  if (good) {
    cout << (n / 2) + 1;
    return 0;
  }
  cout << 1;
  return 0;
}
