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 com.company;

import java.util.HashSet;
import java.util.Scanner;

public class Equation {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        long n = input.nextLong();
        if(n==1l)
            System.out.println(9+" "+8);
        else if(n==2l)
            System.out.println(6+" "+4);
        else if(n==3l)
            System.out.println(9+" "+6);
        else if(n==5l)
            System.out.println(20+" "+15);
        else if(n==6l)
            System.out.println(12+" "+6);
        else if(n==7l)
            System.out.println(21+" "+14);
        else if(n%2l==0l)
        {
            System.out.println(n+4+" "+4);
        }
        else if(n%2!=0l)
        {
            System.out.println(n+9+" "+9);
        }



    }
}