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;
bool primo(long long p) {
  for (long long i = 2; i * i <= p; i++)
    if (p % i == 0) return false;
  return true;
}
long long primeiro(long long a) {
  for (long long i = 2; i * i <= a; i++)
    if (a % i == 0) return i;
}
int main() {
  long long a, n, steps = 0;
  cin >> n;
  if (primo(n)) {
    cout << 1 << "\n";
    return 0;
  }
  while (n > 0) {
    if (n % 2 == 1) {
      a = primeiro(n);
      n -= a;
      steps++;
    } else {
      steps += n / 2;
      n -= n;
    }
  }
  cout << steps << "\n";
  return 0;
}
