#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 9;
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n == 2) return 1;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n;
  cin >> n;
  if (n % 2 == 0)
    cout << n / 2;
  else {
    if (isPrime(n)) {
      cout << 1;
      return 0;
    }
    for (long long i = 3; i * i <= n; i += 2) {
      if (isPrime(i) == 0) continue;
      if (n % i == 0) {
        cout << (n - i) / 2 + 1;
        return 0;
      }
    }
  }
}

### Prompt

Generate 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;
const long long N = 2e5 + 9;
long long isPrime(long long n) {
  if (n < 2) return 0;
  if (n == 2) return 1;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n;
  cin >> n;
  if (n % 2 == 0)
    cout << n / 2;
  else {
    if (isPrime(n)) {
      cout << 1;
      return 0;
    }
    for (long long i = 3; i * i <= n; i += 2) {
      if (isPrime(i) == 0) continue;
      if (n % i == 0) {
        cout << (n - i) / 2 + 1;
        return 0;
      }
    }
  }
}

```