Let s(x) be sum of digits in decimal representation of positive integer x. Given two integers n and m, find some positive integers a and b such that 

  * s(a) ≥ n, 
  * s(b) ≥ n, 
  * s(a + b) ≤ m. 

Input

The only line of input contain two integers n and m (1 ≤ n, m ≤ 1129).

Output

Print two lines, one for decimal representation of a and one for decimal representation of b. Both numbers must not contain leading zeros and must have length no more than 2230.

Examples

Input

6 5


Output

6 
7


Input

8 16


Output

35 
53

Note

In the first sample, we have n = 6 and m = 5. One valid solution is a = 6, b = 7. Indeed, we have s(a) = 6 ≥ n and s(b) = 7 ≥ n, and also s(a + b) = s(13) = 4 ≤ m.
import java.io.*;
import java.math.BigInteger;
import java.util.*;

/**
 * @author gamezovladislav
 */

public class TaskB {
    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Solve solve = new Solve(in, out);
        solve.solve();
        out.close();
    }

    private static class Solve {

        private final InputReader in;
        private final PrintWriter out;

        public Solve(InputReader in, PrintWriter out) {
            this.in = in;
            this.out = out;
        }

        public void solve() {
            int y = 2000;
            for (int i = 0; i < y; i++) {
                out.print("4");
            }
            out.println();
            for (int i = 1; i < y; i++) {
                out.print("5");
            }out.print("6");
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}