#include <bits/stdc++.h>
using namespace std;
bool IsPrime(long long int value) {
  for (int i = 2; i <= sqrt(value); i++)
    if ((value % i) == 0) return 0;
  return 1;
}
int main() {
  bool m = false;
  long long k, l;
  cin >> k;
  for (int i = 2; i <= sqrt(k); i++) {
    if (IsPrime(i) && (k % i == 0)) {
      l = i;
      m = true;
      break;
    }
  }
  if (!m) {
    l = k;
  }
  cout << 1 + (k - l) / 2;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
bool IsPrime(long long int value) {
  for (int i = 2; i <= sqrt(value); i++)
    if ((value % i) == 0) return 0;
  return 1;
}
int main() {
  bool m = false;
  long long k, l;
  cin >> k;
  for (int i = 2; i <= sqrt(k); i++) {
    if (IsPrime(i) && (k % i == 0)) {
      l = i;
      m = true;
      break;
    }
  }
  if (!m) {
    l = k;
  }
  cout << 1 + (k - l) / 2;
  return 0;
}

```