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.util.Scanner;

public class eqaution {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		long n=scn.nextLong();
		long a=0;
		long b=0;
		for(long i=4;;i++) {
			boolean isi=iscomp(i);
			if(isi) {
				boolean isn=iscomp(i+n);
				if(isn) {
					a=i+n ;
					b=i;
					break;
				}
			}
			
		}
		System.out.println(a+" "+b);
		

	}
	public static boolean iscomp(long n) {
		
		
		while(n%2==0) {
			
			n=n/2;
			return true;
		}
		
		for(long i=3;i<=Math.sqrt(n);i+=2) {
			
			while(n%i==0) {
				n=n/i;
				return true;
			}
		}
		return false;
		
	}

}