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.*;
import java.math.*;
import javax.swing.text.html.parser.Entity;

// Itreator for map  for(Map.Entry ent :mp.entrySet())
public class Main {

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        if(n>=1000)
        {
            System.out.println((n+n+n)+" "+(n+n));
        }
        else
        {
        int ind1 = 0, ind2 = 0;
        Set<Integer> st = new HashSet<Integer>();
        int ar[] = new int[1000000], l = 0;
        for (int i = n; i <= 1000000; i += n) {
            for (int k = 2; k <= 9; k++) {
                if (i != k && i % k == 0 && i != n) {
                    st.add(i);
                }
            }
        }
        Iterator it = st.iterator();
        while (it.hasNext()) {
            int cnt = (int) (it.next());
            ar[l] = cnt;
            l++;
            //System.out.println(cnt); 
        }
        Arrays.sort(ar, 0, l);
        for (int i = 0; i < l; i++) {
            if (Math.abs(ar[i] - ar[i + 1]) == n) {
                System.out.println(ar[i+1] + " " + ar[i]);
                break;
            }
        }
        }
        //   System.out.println(ind1+" "+ind2); 
    }

}