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.ArrayList;
import java.util.Collections;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int n=s.nextInt();
		
		for(int i=n+1;i<=1000000000;i++)
		{
			if(prime(i)==false&&prime(i+n)==false)
			{
				System.out.println((i+n)+" "+(i));
				break;
			}
		}
		
	}
	
	public static boolean prime(int a)
	{
		int p=0;
		
		for(int i=2;i<=Math.sqrt(a);i++)
		{
			if(a%i==0)
			{
				p=1;
				break;
			}
		}
		
		if(p==0)
		{
			return true;
		}
		else
			return false;
	}
}