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.*;
public class Main {
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		boolean bb=true;
		for(int i=4;i<=1000000;i++){
			int a=n+i,b=i;
			for(int j=2;j<a;j++){
				if(a%j==0){
					for(int k=2;k<b;k++){
						if(b%k==0){
							System.out.println(Math.max(a,b)+" "+Math.min(a, b));
							bb=false;
							break;
						}
					}
					break;
				}
			}
			if(!bb)
				break;
		}
	}
}