#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<long long> v;
  long long a, z;
  cin >> a;
  z = a;
  for (int i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) {
      z = i;
      break;
    }
  }
  if (a == z)
    cout << 1;
  else if (z == 2)
    cout << a / 2;
  else
    cout << (a - z) / 2 + 1;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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() {
  vector<long long> v;
  long long a, z;
  cin >> a;
  z = a;
  for (int i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) {
      z = i;
      break;
    }
  }
  if (a == z)
    cout << 1;
  else if (z == 2)
    cout << a / 2;
  else
    cout << (a - z) / 2 + 1;
}

```