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 Ishu
{
	public static void main(String[] args)
	{
	Scanner scan=new Scanner(System.in);
	int n,b=4;
	n=scan.nextInt();
	int a=n+4;
	while(b<=990000000)
		{
		a=n+b;
		int temp=0;
		for(int i=2;i*i<=a;++i)
			if(a%i==0)
				{
				++temp;
				break;
				}
		for(int i=2;i*i<=b;++i)
			if(b%i==0)
				{
				++temp;
				break;
				}
		if(temp==2)
			break;
		++b;
		}
	System.out.println(a+" "+b);
	}
}