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 final class CountingInversions {
	public static void main (String[] args) {
	    Scanner in=new Scanner(System.in);
	    int n=in.nextInt();
	    boolean found=false;
	    int x=2;
	    while(!found){
	        if(isNotPrime(x)){
	            if(isNotPrime(x+n)){
	                found=true;
	                System.out.println((x+n)+" "+x);
	                return;
	            }
	        }
	        x++;
	    }
		
	}
	public static boolean isNotPrime(int x){
	    for(int i=2;i<x/2;i++){
	        if(x%i==0)
	        return true;
	    }
	    return false;
	}
}