#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n;
  cin >> n;
  long long int temp = -1;
  if (n % 2 == 0) {
    temp = 2;
  } else if (n % 3 == 0) {
    temp = 3;
  } else {
    for (long long int z = 5; z * z <= n; z += 6) {
      if (n % z == 0) {
        temp = z;
        break;
      } else if (n % (z + 2) == 0) {
        temp = z + 2;
        break;
      }
    }
  }
  if (temp == -1) {
    temp = n;
  }
  long long int p2 = n - temp;
  long long int p3 = p2 / 2;
  cout << p3 + 1 << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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() {
  long long int n;
  cin >> n;
  long long int temp = -1;
  if (n % 2 == 0) {
    temp = 2;
  } else if (n % 3 == 0) {
    temp = 3;
  } else {
    for (long long int z = 5; z * z <= n; z += 6) {
      if (n % z == 0) {
        temp = z;
        break;
      } else if (n % (z + 2) == 0) {
        temp = z + 2;
        break;
      }
    }
  }
  if (temp == -1) {
    temp = n;
  }
  long long int p2 = n - temp;
  long long int p3 = p2 / 2;
  cout << p3 + 1 << endl;
  return 0;
}

```