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 int isComposite(long  x)
    {
        int result = 0;
       long k = 2;
        while(k < x)
        {
          if(x%k == 0)
          {
              result = 1;
              break;
          }
          k = k+1;
        }
        return result;
    }
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
       long  i = 2;
       while(i++ < 90000)
       {
         
        int ans =  isComposite(i);
        if(ans == 1)
        {
          ans =   isComposite(i+n);
          if(ans == 1)
          {
              System.out.println((i+n) + " " + i);
              break;
          }
            
        }
         
       }
        
    }
}