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;
import java.io.PrintWriter;
import java.util.Comparator;
import java.util.List;

public class Equation { // Template for CF
    public static class ListComparator implements Comparator<List<Integer>> {

        @Override
        public int compare(List<Integer> l1, List<Integer> l2) {
            for (int i = 0; i < l1.size(); ++i) {
                if (l1.get(i).compareTo(l2.get(i)) != 0) {
                    return l1.get(i).compareTo(l2.get(i));
                }
            }
            return 0;
        }
    }

    public static void main(String[] args) throws IOException {
        // Check for int overflow!!!!
        // Should you use a long to store the sum or smthn?
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int N = Integer.parseInt(f.readLine());
        out.println((N * 9) + " " + (N * 8));
        out.close();
    }

}