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 A {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		
		int non_primes[] = new int[10000005];
		
		non_primes[0] = non_primes[1] = 0;
		non_primes[2]  = 0;
		for (int i = 2; i < non_primes.length; i++) {
			if(non_primes[i]==0){
				for (int j = 2; i*j < non_primes.length; j++) {
					non_primes[i*j] = 1;
				}
			}
		}
		
		for (int i = 3; i < 10000; i++) {
			
			if(non_primes[i]==1 && non_primes[i+n]==1){
				
				System.out.println((n+i)+" "+i);
				return;
			}
		}
	}

}