#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  long long min_d = 0;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      min_d = i;
      break;
    }
  }
  if (min_d == 0) {
    min_d = x;
  }
  long long result;
  if (min_d == 2) {
    result = x / 2LL;
  } else {
    result = 1 + (min_d * (x / min_d - 1LL)) / 2;
  }
  cout << result << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long x;
  cin >> x;
  long long min_d = 0;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      min_d = i;
      break;
    }
  }
  if (min_d == 0) {
    min_d = x;
  }
  long long result;
  if (min_d == 2) {
    result = x / 2LL;
  } else {
    result = 1 + (min_d * (x / min_d - 1LL)) / 2;
  }
  cout << result << endl;
  return 0;
}

```