#include <bits/stdc++.h>
using namespace std;
long long prime(long long n) {
  if (n == 2 || n == 3) return 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << (n / 2) << endl;
  } else {
    if (n % 3 == 0) {
      long long k = n - 3;
      cout << k / 2 + 1 << endl;
    } else if (prime(n))
      cout << "1" << endl;
    else {
      for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) {
          long long c = n - i;
          cout << (1 + c / 2) << endl;
          break;
        }
      }
    }
  }
  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;
long long prime(long long n) {
  if (n == 2 || n == 3) return 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long n;
  cin >> n;
  if (n % 2 == 0) {
    cout << (n / 2) << endl;
  } else {
    if (n % 3 == 0) {
      long long k = n - 3;
      cout << k / 2 + 1 << endl;
    } else if (prime(n))
      cout << "1" << endl;
    else {
      for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) {
          long long c = n - i;
          cout << (1 + c / 2) << endl;
          break;
        }
      }
    }
  }
  return 0;
}

```