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 first_prime(int num) {
  if (num % 2 == 0) return (2);
  for (int i = 3; i < sqrt(num); i += 2) {
    if (num % i == 0) return (i);
  }
  return (num);
}
int main() {
  int n, x, d = 0, numbers[100] = {0}, counter = 0;
  bool smallest = true;
  cin >> n;
  while (n != 0) {
    x = first_prime(n);
    n = n - x;
    counter++;
  }
  cout << counter;
  return 0;
}
