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.*;
public class WrongSubtraction {
    public static boolean composite(int n) {
        if(n==2) return false;
        for(int i = 2; i <= (int)Math.sqrt(n);++i) {
            if(n%i == 0)
                return true;
        }
        return false;
    }
	public static void main(String args[]) throws IOException {
	    
	    // input
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine()); // 1
		int i = 2, a = -99, b = n;
		do { // 0 != 2
		    b = b*(int)Math.pow(2, ++i);
		    if(composite(b)){ // true
		        a = b+n; // 9
		        if(composite(a)) {
		            System.out.println(a+" "+b);
		            break;
		        } else a = -99;
		    }
		} while(a != b+n);
	}
}