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;

public class Equation {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n, a, b,c=0,c1=0;
        n = input.nextInt();
          a = n + 1;
        

        while (true) {
           
            for (int i = 2; i <= a / 2; ++i) {

                if (a % i == 0) {
                    c++;
                    break;
                }
            }
              b = a - n;
          if(c!=0){
          

            for (int i = 2; i <= b / 2; ++i) {
                
                if (b % i == 0) {
                    c1++;
                    break;
                }
            }
          }
            
            if(c!=0 && c1!=0){
                System.out.print(a+" ");
                  System.out.print(b);
                  break;
            }else{
                a++;
                c=0;
                c1=0;
                continue;
            }

        }
    }

}