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 IsPrime(long long);
int main() {
  long long N, D = 2, Answer = 0;
  cin >> N;
  if (IsPrime(N)) D = N;
  while (N % D != 0 || !IsPrime(D)) ++D;
  cout << N / D;
  return 0;
}
bool IsPrime(long long n) {
  long k = sqrt(n);
  for (long i = 2; i <= k; ++i)
    if (n % i == 0) return false;
  return true;
}
