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 count(long long n) {
  if (n % 2 == 0)
    return 2;
  else {
    for (long long i = 3; i <= sqrt(n); i += 2) {
      if (n % i == 0) {
        return i;
      }
    }
    return n;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n;
  cin >> n;
  long long x;
  if (n % 2 == 0)
    cout << n / 2 << endl;
  else {
    int f = 0;
    int c = 0;
    while (n > 0 && count(n) > 2) {
      n = n - count(n);
      c++;
    }
    c += n / 2;
    cout << c << endl;
  }
}
