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 massa {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int a = 1, b = 1, n;
		n = in.nextInt();
		a = a + n;
		int ContB = 2;
		int ContA=2;
		while (ContA != -1) {
			if (a % ContA == 0) {
				
				while (ContB != -1) {
					if (b % ContB == 0 && ContB<b) {
						ContB = -1;
						ContA=-1;
					}else if(ContB>= b/2) {
						ContA++;
						ContB=2;
						break;
					}
					else if (b % ContB != 0) {
						ContB++;
					} else {
						break;
					}

				}

			} else if (ContA <= a / 2 && a % ContA != 0) {
				ContA++;
			}
			else {
				a++;
				b++;
				ContA=2;
			}

		}
System.out.println(a);
System.out.println(b);
	}

}