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 HHH(int a) {
        boolean d=false;
        for (int i=2;i<=Math.sqrt(a);i++) {
            if (a%i==0) {
                d=true;
                break;
            }
        }
        if (d==false) return true;
        else return false;
    }
    public static void main(String[] args) {
        Scanner scanner = new Scanner (System.in);
        int n = scanner.nextInt();
        for (int i=4;i<1000000000;i++) {
            if (HHH(i)==false&&HHH(i+n)==false) {
                System.out.print((i+n) + " " + i);
                break;
            }
        }

        scanner.close();
    }
}