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.io.PrintWriter;
import java.util.Scanner;

public class EquationA extends PrintWriter {
    //this trick improves performances
    EquationA() {
        super(System.out);
    }

    public static void main(String[] $) {
        EquationA o = new EquationA();
        o.main();
        o.flush();
    }
    int[] primes = {2,5,7,9,11,13,17,23,29,31,37};

    void main() {
        Scanner sc = new Scanner(System.in);
        //int count = sc.nextInt();
        //use this if just a single test
        int count = 1;
        ext: while (count-- > 0) {
            final int n = sc.nextInt();
     
            int tmp=0;
            for(int j=2;j<Integer.MAX_VALUE;j++){
                tmp=2*j+n;
                for(int div:primes){
                    if(tmp%div==0&&tmp/div>1){
                        println((2*j+n)+" "+(2*j));
                        continue ext;
                    }
                }
            }
            println("0 0");
        }
        sc.close();
    }
}