#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  }
  long long int d;
  d = n;
  long long int sq = sqrt(n);
  for (long long int i = 3; i <= sq; i++) {
    if (n % i == 0) {
      d = i;
      n = ((n - d) / 2) + 1;
      cout << n;
      return 0;
    }
  }
  n = ((n - d) / 2) + 1;
  cout << n;
  return 0;
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2;
    return 0;
  }
  long long int d;
  d = n;
  long long int sq = sqrt(n);
  for (long long int i = 3; i <= sq; i++) {
    if (n % i == 0) {
      d = i;
      n = ((n - d) / 2) + 1;
      cout << n;
      return 0;
    }
  }
  n = ((n - d) / 2) + 1;
  cout << n;
  return 0;
}

```