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;
int isprime(long long int n) {
  if (n == 1) {
    return (0);
  }
  if (n == 2) {
    return (1);
  }
  for (long long int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return (0);
    }
  }
  return (1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int f = 0, m = INT_MAX;
  for (long long int i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      if (isprime(i)) {
        m = min(m, i);
        f = 1;
      }
      if (isprime(n / i)) {
        m = min(m, n / i);
        f = 1;
      }
    }
  }
  if (f == 0) {
    cout << "1";
  } else {
    long long int ans = n / m;
    cout << ans;
  }
  return (0);
}
