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.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class test {
	public static class pair implements Comparable<pair> {
		int t1 = 0, t2 = 0;

		public pair(int t1) {
			this.t1 = t1;
		}

		@Override
		public int compareTo(pair o) {
			// TODO Auto-generated method stub
			return this.t2 - o.t2;
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
//		int m = sc.nextInt();
//		StringBuilder st = new StringBuilder();
//		while (m-- > 0) {
//
//			
//		}
//		System.out.println(st);

		int n=sc.nextInt();
		int a=0,b=0;
		if(n%2==0) {
			b=6;
			a=n+b;
		}
		else {
			b=9;
			a=n+b;
		}
		System.out.println(a+" "+b);
		
	}

}