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 n, subs, first, p, pr;
int main() {
  cin >> n;
  for (int d = 2; d <= sqrt(n); d++) {
    if (n % d == 0) {
      if (first == 0) first = d;
      break;
    }
  }
  if (first == 0)
    cout << 1;
  else {
    pr = n / first;
    if (first == 2)
      cout << pr;
    else
      cout << 1 + (n - first) / 2;
  }
  return 0;
}
