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 func(long long int n) {
  long long int count = 0, i;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) count++;
    long long int k = n / i;
    if (k != i && n % k == 0) count++;
  }
  if (count <= 2) return 1;
  return 0;
}
int main() {
  long long int n, i, j, count = 0;
  cin >> n;
  set<long long int> s;
  for (i = 1; i <= sqrt(n); i++) {
    if (i != 1 && n % i == 0) {
      s.insert(i);
      break;
    }
    long long int k = n / i;
    if (k != 1 && n % k == 0) {
      s.insert(k);
    }
  }
  auto itr = s.begin();
  cout << 1 + (n - (*itr)) / 2 << endl;
}
