#include <bits/stdc++.h>
using namespace std;
long long int div(long long int n) {
  int top = sqrt(n);
  for (int i = 2; i <= top; i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return n;
}
int main() {
  long long int n;
  cin >> n;
  int c = 0;
  while (n % 2 != 0) {
    n -= div(n);
    c++;
  }
  cout << c + n / 2;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
long long int div(long long int n) {
  int top = sqrt(n);
  for (int i = 2; i <= top; i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return n;
}
int main() {
  long long int n;
  cin >> n;
  int c = 0;
  while (n % 2 != 0) {
    n -= div(n);
    c++;
  }
  cout << c + n / 2;
  return 0;
}

```