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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class C {

	public static void main(String[] args) {
		try {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            String linea;
            linea = br.readLine();
            int n = Integer.parseInt(linea);
            int a = 0; int b = 0;
            boolean bb = false;
            int i = 7;
            int c = 0, cb = 0;
            while(bb==false) {
            	bb=false;
            	c=0;
            	cb= 0;
            	i++;
            	a=i;
            	b=a+n;
            	for(int j = 1; j <= a; j++)
            	{
            	    if((a % j) == 0){
            	        c++;
            	    }
            	}
            	for(int j = 1; j <= b; j++)
            	{
            	    if((b % j) == 0){
            	        cb++;
            	    }
            	}

            	if(c > 2 && cb > 2)
            	{
            		bb=true;
            	}
            	
            }

            System.out.println(b + " " + a);

            
	} catch (IOException e) {
	    e.printStackTrace();
	}


	}
}