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.CompititivePrograming;

// Code Contributed by Mohit Gupta_OMG <(0-o)>
// Java program to demonstrate working of extended
// Euclidean Algorithm

import java.util.Scanner;

public class gcd
{
    // extended Euclidean Algorithm
   static  int gcdM(int a, int b, int x, int y,A obj) {
        if (b == 0) {
            obj.x = 1;
            obj.y = 0;
            return a;
        }
        int x1=1, y1=1;
        int d = gcdM(b, a % b, x1, y1,obj);
        obj.x =-y1;

        obj.y = x1 + y1 * (a / b);;
        return d;
    }

   static class A{
        static int x=1;
        static  int y=1;
    }
    // Driver Program
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);
        int a = 3, b = 2;
        int n=sc.nextInt();
        A obj=new A();
        int g = gcdM(a, b, 1,1,obj);
        int x0=obj.x;
        int y0=obj.y;
        int c=n;
        x0 *= c / g;
        y0 *= c / g;
        //if (a < 0) x0 = -x0;
        //if (b < 0) y0 = -y0;
        System.out.print(9*n+" "+8*n);

    }
}
// Code Contributed by Mohit Gupta_OMG <(0-o)>