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[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int a = 4;
		int b = a + n;
		while(true) {
			int flag = 0;
			for(int i = 2; i<b/2; i++) {
				if(b%i == 0) {
					flag++;
					break;
				}				
			} 
			for(int i = 2; i<a/2; i++) {
				if(a%i == 0) {
					flag++;
					break;
				}				
			}
			if(flag == 2) 
				break;		
			a++;
			b++;
		}
		System.out.println(b + " " + a);	
		sc.close();
	}
}