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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class TaskA {
	public static void main(String[] arg) {
		final FastScanner in = new FastScanner(System.in);
		final PrintWriter out = new PrintWriter(System.out);
		final int n = in.nextInt();
		int b = 4;
		for (; b <= 100 * n; b++) {
			if (!isPrime(b) && !isPrime(b + n)) {
				break;
			}
		}
		out.printf("%d %d%n", b+n, b);
		out.close();
		in.close();
	}

	private static boolean isPrime(int n) {
		if (n == 1) {
			return true;
		}
		if (n == 2) {
			return true;
		}
		if (n == 3) {
			return true;
		}
		for (int i = 2; i * i <= n; i++) {
			if (n % i == 0)
				return false;
		}
		return true;
	}

	private static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		FastScanner(InputStream stream) {
			try {
				br = new BufferedReader(new InputStreamReader(stream));
			} catch (Exception e) {
				e.printStackTrace();
			}
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		int[] readIntArr(int n) {
			int[] result = new int[n];
			for (int i = 0; i < n; i++) {
				result[i] = Integer.parseInt(next());
			}
			return result;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}

		long nextLong() {
			return Long.parseLong(next());
		}

	}

}