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