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.
	import java.util.*;
import java.math.*;	
	public class Main {
		public static boolean isPrime(long number) {
	        int sqrt = (int) Math.sqrt(number) + 1;
	        for (int i = 2; i < sqrt; i++) {
	            if (number % i == 0) {
	                // number is perfectly divisible - no prime
	                return false;
	            }
	        }
	        return true;
	    }


	//Read more: http://www.java67.com/2014/01/how-to-check-if-given-number-is-prime.html#ixzz5WebGihHR



	//Read more: http://www.java67.com/2014/01/how-to-check-if-given-number-is-prime.html#ixzz5WebGihHR

		static final int MAX=1000000007 ;
	    public static void main(String[] args) {
			Scanner s=new Scanner(System.in);
			long n=s.nextLong();
			int cnt=0;
			if(isPrime(n)) {
				System.out.println(1);
			}
			else {
				if((n%2)==0) {
					System.out.println((n/2));
				}
				else {
					n=n-1;
					System.out.println((n/2));
				}
			}
			
			
			
			
	    }
	}