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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, flag = 0;
  cin >> n;
  vector<bool> v(100001, true);
  for (int i = 2; i * i <= 100000; i++) {
    if (v[i] == true) {
      for (int j = i * i; j <= 100000; j += i) v[j] = false;
    }
  }
  if (n == 0)
    cout << 0 << endl;
  else if (!(n & 1))
    cout << n / 2 << endl;
  else {
    for (int i = 2; i <= 100000; i++)
      if (v[i] == true)
        if (n % i == 0) {
          cout << (n - i) / 2 + 1 << endl;
          flag = 1;
          break;
        }
    if (flag == 0) cout << 1 << endl;
  }
}
