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 int smallest_prime_divisor(long long int x) {
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      return i;
    }
  }
  return x;
}
void solve() {
  long long int n, count = 0;
  cin >> n;
  while (n % 2 != 0) {
    long long int temp = smallest_prime_divisor(n);
    n -= temp;
    count++;
  }
  cout << (count + (n / 2));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) solve();
  return 0;
}
