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.util.ArrayList;
public class Main{   
   
   
    public static void main(String[] args){
     Scanner param = new Scanner(System.in);
		long n=param.nextLong();
	   if(n%2==0){
	      
	      System.out.print(n+4+" ");
	      System.out.print(4);
	      return ;
	   }
	   if(n%2!=0){
	       if((n+2)%2==0){
	            
	            System.out.print(n+4+" ");
	            System.out.print(4);
	            return ;
	       }
	       if((n+3)%2==0){
	          
	            System.out.print(n+9+" ");
	            System.out.print(9);
	            return ;
	       }
	   }
}
}