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.Scanner;
public class nha_toan_hoc_cuc_nhanh {
public static boolean kiemtra(long a) {
	if(a<2) {
		return false;
	}
		for(int i=2; i<=Math.sqrt(a); i++) {
			if(a%i==0) {
				return false;// k phai nguyen to
			}
		}
		return true;
	}
	
	public static void main(String[] args) {
		Scanner SC = new Scanner(System.in);
		long n;
		n=SC.nextLong();
		long a=0, b=99999999;	
		boolean f=false;
		while(f==false){
			
			a=b-n;
			boolean g1=kiemtra(a);
			boolean g2=kiemtra(b);
			if(g1 == false && g2 == false ) {
				f=true;
			}else {
				b--;
			}	
		}
		System.out.println(b);
		System.out.println(a);
		SC.close();
}
}