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 primeFactors(int n) {
  while (n % 2 == 0) {
    return 2;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      return i;
      n = n / i;
    }
  }
  if (n > 2) return n;
}
int main() {
  int x;
  cin >> x;
  int cnt = 0;
  while (x) {
    int ans = primeFactors(x);
    cnt++;
    x -= ans;
  }
  cout << cnt;
}
