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 AOddSum {

	public static void main(String[] args) {
		
		Scanner scn = new Scanner(System.in);
		int n = scn.nextInt();
		int z = n+4;
		int y = n+9;
		int a1 = Math.abs(n-4);
		int a2 = Math.abs(n-9);
		if(a1%2==0) {
			System.out.println(z + " " + 4);
		}
		else {
			System.out.println(y + " " + 9);
		}
		
	}		
}