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 int getsmall(long long int x) {
  for (long long int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return i;
  }
  return x;
}
int main() {
  long long int n;
  long long int count = 0;
  cin >> n;
  if (n == 0) return 0;
  if (n % 2 != 0) {
    n = n - getsmall(n);
    count++;
  } else
    count = count + (n / 2);
  cout << count << endl;
  return 0;
}
