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 isPrime(long long number) {
  if (number < 2) return false;
  if (number == 2) return true;
  if (number % 2 == 0) return false;
  for (long long i = 3; (i * i) <= number; i += 2) {
    if (number % i == 0) return false;
  }
  return true;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  long long ans = 0;
  while (n != 0) {
    if (n % 2 == 0) {
      ans += n / 2;
      break;
    } else if (isPrime(n)) {
      n = 0;
      ans++;
    } else {
      for (long long i = 2; i * i <= n; ++i) {
        if (n % i == 0) {
          if (isPrime(n / i) && isPrime(i))
            n -= min(n / i, i), ans++;
          else if (isPrime(i))
            n -= i, ans++;
          else if (isPrime(n / i))
            n -= n / i, ans++;
        }
      }
    }
  }
  cout << ans << '\n';
  return 0;
}
