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.*;
public class DiviDiff {
		static Scanner in=new Scanner(System.in);
		static long n=in.nextLong();
		static boolean prime[] = new boolean[100000]; 
	    public static void sieveOfEratosthenes(long n){
	        for(int i=0;i<100000;i++) 
	            prime[i] = true;
	        for(int p = 2; p <100000; p++){
	            if(prime[p] == true) { 
	                for(int i = p*2; i<100000; i += p) 
	                    prime[i] = false; 
	            } 
	        }
	    }
	    static boolean isPrime(long n) {
	        if (n <= 1) return false; 
	        if (n <= 3) return true;
	        if (n % 2 == 0 || n % 3 == 0) return false; 
	        for (int i = 5; i * i <= n; i = i + 6) 
	            if (n % i == 0 || n % (i + 2) == 0) 
	            return false;
	        return true; 
	    } 
	public static void main(String[] args) {
		sieveOfEratosthenes(n);
		long count=0,temp=n;
		 if(isPrime(n))
         	System.out.println("1");
         else if(n%2==0) System.out.println(n/2);
	    else {
	    	for(int p = 2; p*p <=n; p++){
	            if(prime[p] == true) { 
	            	count=p;break;
	            }
	        }
	    	System.out.println(((n-count)/2)+1);
	    }
	}
}