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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  fast();
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long k = n;
    long long i = 2;
    while (i <= sqrt(n) + 1) {
      if (n % i == 0) {
        k = i;
        break;
      }
      i += 1;
    }
    if (k == n)
      cout << 1;
    else
      cout << n / k;
  }
  return 0;
}
