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.
#include <bits/stdc++.h>
using namespace std;
long long n;
bool Check(long long num) {
  if (num == 1) return false;
  if (num == 2 || num == 3) {
    return true;
  }
  if (num % 6 != 1 && num % 6 != 5) {
    return false;
  }
  for (long long i = 5; i * i <= num; i += 6) {
    if (num % i == 0 || num % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
int main() {
  cin >> n;
  if (n % 2 == 0) {
    cout << n / 2 << endl;
  } else {
    long long cnt = 0;
    long long ans;
    if (Check(n)) {
      cout << "1" << endl;
    } else {
      while (n) {
        long long i;
        for (i = 2; i <= n; i++) {
          if (n % i == 0) {
            break;
          }
        }
        n -= i;
        cnt++;
        if (Check(n)) {
          cout << cnt + 1 << endl;
          break;
        }
        if (n % 2 == 0) {
          cout << cnt + n / 2 << endl;
          break;
        }
      }
    }
  }
  return 0;
}
