#include <bits/stdc++.h>
using namespace std;
long long spd(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return n;
}
int main() {
  long long k;
  cin >> k;
  if (k % 2 == 0) {
    cout << (k / 2);
  } else {
    long long d = spd(k);
    if (k == d) {
      cout << (1);
    } else {
      k = k - d;
      cout << ((k / 2) + 1);
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 spd(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return i;
    }
  }
  return n;
}
int main() {
  long long k;
  cin >> k;
  if (k % 2 == 0) {
    cout << (k / 2);
  } else {
    long long d = spd(k);
    if (k == d) {
      cout << (1);
    } else {
      k = k - d;
      cout << ((k / 2) + 1);
    }
  }
  return 0;
}

```