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;
bool prime(long long n) {
  long long a = sqrt(n);
  for (long long i = 2; i <= a; i++)
    if (n % i == 0) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  vector<long long> v;
  long long s = 0;
  for (long long i = 2; i < 100000; i++) {
    if (prime(i)) {
      v.push_back(i);
    }
  }
  s = v.size();
  long long n, ans = 0;
  cin >> n;
  for (long long i = 0; i < s; i++) {
    if (n == 0) break;
    if (n % v[i] == 0) {
      ans++;
      n -= v[i];
      ans += n / 2;
      break;
    }
    if (n == 0) break;
  }
  cout << ans;
  return 0;
}
