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 int k) {
  if (k % 2 == 0) {
    return 1;
  }
  for (long long int i = 2; i < sqrt(k); ++i) {
    if (k % i == 0) {
      return 1;
    }
  }
  return 0;
}
long long int getMinPrime(long long int k) {
  long long int ans;
  if (k % k == 0)
    if (prime(k) == 0) return (k);
  for (long long int i = sqrt(k); i >= 2; --i)
    if (k % i == 0)
      if (prime(i) == 0) return i;
  return 2;
}
int main(int argc, const char* argv[]) {
  {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  }
  cout << "k" << endl;
  long long int k;
  cin >> k;
  long long int cnt = 0;
  while (k > 0) {
    ++cnt;
    k -= getMinPrime(k);
  }
  cout << "cnt: " << cnt << endl;
  return 0;
}
