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.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        boolean[] crossed = new boolean[100000001];
        for (int i = 2; i < 20000001; i++) {
            if (!crossed[i]) {
                for (int j = i+i; j < 100000001; j += i) {
                    crossed[j] = true;
                }
            }
        }
        int n = Integer.parseInt(in.readLine());
        for (int i = n; i < 100000001; i++) {
            int b = i-n;
            if (crossed[i] && crossed[b]) {
                System.out.println(i + " " + b);
                return;
            }
        }
    }

}
/*


 */