#include <bits/stdc++.h>
using namespace std;
bool judge(long long x) {
  long long z = sqrt(x);
  for (long long i = 2; i <= z; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long x;
  scanf("%lld", &x);
  if (judge(x))
    printf("1\n");
  else {
    if (x % 2 == 0)
      printf("%lld\n", x / 2);
    else {
      long long z = sqrt(x);
      for (long long i = 2; i <= z; i++) {
        if (x % i == 0 && judge(i)) {
          x -= i;
          break;
        }
      }
      printf("%lld\n", x / 2 + 1);
    }
  }
  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 judge(long long x) {
  long long z = sqrt(x);
  for (long long i = 2; i <= z; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  long long x;
  scanf("%lld", &x);
  if (judge(x))
    printf("1\n");
  else {
    if (x % 2 == 0)
      printf("%lld\n", x / 2);
    else {
      long long z = sqrt(x);
      for (long long i = 2; i <= z; i++) {
        if (x % i == 0 && judge(i)) {
          x -= i;
          break;
        }
      }
      printf("%lld\n", x / 2 + 1);
    }
  }
  return 0;
}

```