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:
/******************************************************************************

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Write your code in this editor and press "Run" button to execute it.

*******************************************************************************/
import java.util.Scanner;
public class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int a=n+2;
		int b=2;
        int found=0;
        while(found==0)
        {
            int ca=0,cb=0;
           for(int i=2;i<=a;i++)
           {
               if(a%i==0 && a!=i)
               {
                   ca++;
               }
           }
           for(int i=2;i<=b;i++)
           {
               if(b%i==0 && b!=i)
               {
                   cb++;
               }
           }
           
           if( ca>0 && cb>0)
           {
               found=1;
           }
           else
           {
               ca=0;
               cb=0;
               a++;
               b++;
           }
        }
        System.out.println(a+" "+b);
		
	}
}