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;
int pp(long long int n, long long int c) {
  if (n == 0)
    return c;
  else if (n % 2 == 0) {
    c = c + n / 2;
    return c;
  } else if (n % 3 == 0) {
    c = c + n / 3;
    return c;
  } else {
    long long int i;
    for (i = 5; i * i <= n; i = i + 6) {
      if (n % i == 0) return (c + n / i);
      if (n % (i + 2) == 0) return (c + n / (i + 2));
    }
    return 1;
  }
}
int main() {
  long long int n, c, t;
  cin >> n;
  c = 0;
  t = pp(n, c);
  cout << t;
}
