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.io.BufferedReader;
import java.io.InputStreamReader;
 
public class Solution
{
    public static boolean isComposite(int n)
    {
        for(int i = 2; i <= Math.sqrt(n) ; i++)
            if(n % i == 0)
                return true;
        return false;
    }
    public static void main(String[] args)throws Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        for(int i = 8; i < 3 * n + 14; i++)
        {
            if(isComposite(i) && isComposite(i+n))
            {
                System.out.println((i+n) +" "+ i);
                break;
            }
        }
    }
}