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.Scanner;

public class Solution {

	public static void main(String... s) {
	Scanner sc=new Scanner(System.in);
	int a=sc.nextInt();
     if(a%2==0) {
    	System.out.print(a+4+" "+"4");
    	
     }else {
    	 if(a%3==0) {
    		 System.out.print(a+6+" "+"6"); 
    	 }else if((a%3==1||a%3==2)&&a%10==7) {
    		 System.out.print(a+8+" "+"8"); 
    	 }else if((a%3==1||a%3==2)&&a%10==9) {
    		 System.out.print(a+6+" "+"6"); 
    	 }
    	 else if((a%3==1||a%3==2)&&a%10==5) {
    		 System.out.print(a+10+" "+"10"); 
    	 }else if((a%3==1||a%3==2)&&a%10==1) {
    		 System.out.print(a+14+" "+"14"); 
    	 }else if((a%3==1||a%3==2)&&a%10==3) {
    		 System.out.print(a+12+" "+"12"); 
    	 }
     }
	
	}

}