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:
/*package whatever //do not write package name here */

import java.io.*;
import java.util.Scanner;

public class GFG {
    
    public static boolean check(long x){
        for(int i=2;i<x;i++)
        if(x%i==0)
        return true;
    
    return false;
    }
	public static void main (String[] args) {
// 		System.out.println("GfG!");
Scanner sc=new Scanner(System.in);
long n=sc.nextLong();
long temp1=0;
long temp2=0;
for(long i=n;;i++){
    if(check(i) &&check(i+n)){
        temp1=i;
        temp2=i+n;
        break;
    }
}
System.out.println(temp2+" "+temp1);

	}
}