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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

/*
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 * and open the template in the editor.
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/**
 *
 * @author RUTHB
 */
public class dos {

    public static void main(String[] args) throws IOException {
        BufferedReader br
                = new BufferedReader(new InputStreamReader(System.in));
        //int cant = Integer.parseInt(br.readLine());

        long n = Integer.parseInt(br.readLine());
        long a = n + 2;
        // long i = ;
        while (a++ != 0 && a <= 1000000000) {
            long b = a + n;
            //System.out.println(a + " " + b);
            if (factor(a) == true && (factor(b) == true)) {
                System.out.println(b + " " + a);
                a = 0;
            }

        }
    }

    public static boolean factor(long numero) {
        int num = 2;
        //int numeros[] = new int[numero];
        int i = 0;
        while (numero != 1) {

            while (numero % num == 0) {
                //numeros[i] = num;

                if (num != numero) {
                    i++;
                }
                numero /= num;

                if (i != 0) {
                    return true;
                }
            }
            num++;

        }
        if (i != 0) {
            return true;
        } else {
            return false;
        }

    }
}