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>
const int INF = 1 << 29;
int calc(long long n) {
  int ans = n;
  for (int i = 2; i <= n; i++) {
    if (n % i == 0) {
      ans = std::min(ans, i);
    }
  }
  return ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  ;
  long long n;
  std::cin >> n;
  long long mn = calc(n);
  std::cout << '>' << "mn" << ':' << mn << '\n';
  ;
  std::cout << 1 + (n - mn) / 2;
  return 0;
}
