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 comp
{
    public boolean composite(int a)
    {
        for(int j=2;j*j<=a;j++)
        {
            if(a%j==0)
            return true;
            else
            continue;
        }
        return false;
    }
    public static void main(String args[])
    {
        Scanner in= new Scanner(System.in);
        int n=in.nextInt();
        comp ob=new comp();
        for(int i=4;i<=10000000;i+=2)
        {
            if(ob.composite(i+n))
            {
            System.out.println((i+n)+" "+i);
            break;
            }
            else
            continue;
        }
    }
}