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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int n = sc.nextInt();
            if (n == 1) {
                System.out.println(9 + " " + 8);
                continue;
            }
            boolean flag = false;
            for (int i = 2; i * i <= n; i++) if (n % i == 0) flag = true;
            if (flag) System.out.println((n << 1) + " " + n);
            else System.out.println(((n << 1) + n) + " " + (n << 1));
        }
    }
}