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 {
    public static boolean composite(int n) {
        int m = (int) Math.sqrt(n);
        for (int i = 2; i <=m ; i++) {
            if (n % i == 0)
                return true;
        }
        return false;
    }
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int a = 4;
        int b = 4+n;
        while (true) {
            if (composite(a) & composite(b))
            {
                System.out.println(b+ " " + a);
                break;
            }
            a++;
            b++;
        }
    }
}