Problem: Let's call a positive integer composite if it has at least one divisor other than 1 and itself. For example:

  * the following numbers are composite: 1024, 4, 6, 9; 
  * the following numbers are not composite: 13, 1, 2, 3, 37. 



You are given a positive integer n. Find two composite integers a,b such that a-b=n.

It can be proven that solution always exists.

Input

The input contains one integer n (1 ≤ n ≤ 10^7): the given integer.

Output

Print two composite integers a,b (2 ≤ a, b ≤ 10^9, a-b=n).

It can be proven, that solution always exists.

If there are several possible solutions, you can print any. 

Examples

Input


1


Output


9 8


Input


512


Output


4608 4096

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class equation
{
	static Scanner sc =new Scanner(new BufferedReader(new InputStreamReader(System.in)));
	public static void main(String[] args) {
		
		int num=sc.nextInt();
		int n=9*num;
		boolean prime[] = new boolean[n+1]; 
        Arrays.fill(prime,true);
          
        for(int p = 2; p*p <=n; p++) 
        { 
            
            if(prime[p] == true) 
            { 
                // Update all multiples of p 
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 

        for (int i=n; ; i--) {
        	if(prime[i] == false && prime[i-num]==false)
        	{
        		System.out.println(i);
        		System.out.println(i-num);
        		break;
        	}
        	
        	
        }

		
	}
 
	
}