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 {
		static boolean prime[];
		static ArrayList<Integer>pi;
		static void siv(int n)
		{
		    prime=new boolean [n+1];
		    for(int i=0;i<n;i++) {
		    	prime[i]=true;
		    }
		    pi=new ArrayList<Integer>();
		    for (int p=2; p*p<=n; p++)
		    {
		        if (prime[p] == true)
		        {
		            for (int i = p*2; i<=n; i += p)
		            prime[i] = false;
		        }
		    }
		    int x = 1;
		    for(int k = 2; k <= n; k++)
		    {
		        if(prime[k] == true)
		        {
		            pi.add(k);
		            x++;
		        }
		    }
		}

		static final int MAX=1000000007 ;
	    public static void main(String[] args) {
			Scanner s=new Scanner(System.in);
			long n=s.nextLong();
			siv(100005);
			//System.out.println(pi);
			for(int i=0;i<pi.size();i++) {
				//System.out.println(pi.get(i));
				if((n%pi.get(i))==0) {
					if(n%2==0) {
					System.out.println((n/2));
					return;
					}
					else {
						n=n-pi.get(i);
						System.out.println((n/2)+1);
						return;
					}
				}
			}
			System.out.println(1);
			
			
			
	    }
	}