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:
/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

public class GFG {
    static boolean check(int n) 
    { 
        // Corner cases 
        if (n <= 1)  
        return false; 
          
        if (n <= 3)  
        return false ; 
  
        // This is checked so that we can skip  
        // middle five numbers in below loop 
        if (n % 2 == 0 || n % 3 == 0) return true; 
  
        for (int i = 5; i * i <= n; i = i + 6) 
            if (n % i == 0 || n % (i + 2) == 0) 
            return true; 
  
        return false; 
    } 
	public static void main (String[] args) {
	       Scanner s  = new Scanner (System.in);
	       int n  =  s.nextInt();
	       int a  =0;
	       int b =0;
	       
	       for (int i =2; i<=1000000000;i++) {
	            int  k = i+n;
	            if(check(i) && check(k) ) {
	                System.out.println(k+" "+i);
	                break;
	            }
	       }
	      
	       
	}
}