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.ArrayList;
        import java.util.Arrays;
        import java.util.Scanner;

public class Singleton_Pattern {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a= 0;
        int b =0;
        if (n%2==0){
            a = n+4;
            b=4;
        }
        else {
            int x=9;
            for (int i = 1; (int)Math.pow(3,i-1)<n ; i++) {
                x=(int)Math.pow(3,i+1);
            }
            a=x;
            b=x-n;
        }
        System.out.println(a+" "+b);
    }
}