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 Equation {
	private static boolean isComposite(int integer) {
		for (int i = integer-1; i >= 2; i--) {
			if (integer % i == 0) {
				return true;			
			}
		}
		return false;
	}
	public static int[] getCompositeInteger(int integer) {
		int startingA = 4;
		int startingB = 4;
		int difference = integer;
		startingA += difference;
		int[] solution = new int[2];
		solution[0] = startingA;
		solution[1] = startingB;
		while (!isComposite(startingA) || !isComposite(startingB)) {
			startingA++;
			solution[0] = startingA;
			startingB++;
			solution[1] = startingB;
		}
		return solution;
	}
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int number = input.nextInt();
		input.close();
		int[] ans = getCompositeInteger(number);
		System.out.println(ans[0] + " " + ans[1]);
	}
}