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.Scanner;
import java.util.Arrays;
import java.util.*;
public class equation
{
    int isprime(int n)
    {
        int i;
        for(i=2;i<=n/2;i++)
        {
            if(n%i==0)
            return 0;
        }
            return 1;
     }
    public static void main(String args[])
     {
        Scanner s=new Scanner(System.in);
        int n,a=3,b,i; 
        n=s.nextInt();
        b=a+n;
        equation e=new equation();
        for(i=0; ;i++)
        {    
            if(e.isprime(a)==0&&e.isprime(b)==0)
            {
                if(b-a==n){
                System.out.print(b+" "+a);break;}
                else
                a++;b++;
            }
            else
            a++;b++;
        }
    }
}