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.*;
public class Main
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		long n=sc.nextLong();
		long a1=0,b1=0;
		for(long a=n+1;a<=1000000000;a++){
		    boolean c=false;
		    boolean end=false;
		    for(long j=2;j<=Math.sqrt(a);j++){
		        if(a%j==0){
		            c=true;
		            break;
		        }
		    }
		    if(c){
		        long b=a-n;
		        for(long k=2;k<=Math.sqrt(b);k++){
		            if(b%k==0){
		                a1=a;
		                b1=b;
		                end=true;
		                break;
		            }
		        }
		    }
		    if(end){
		        break;
		    }
		}
		System.out.println(a1+" "+b1);
	}
}