Problem: 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.

Solution (in JAVA):

Solution:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class AIMRound5B {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		out=new PrintWriter (new BufferedOutputStream(System.out));
		FastReader s=new FastReader();
		int n=s.nextInt();
		int m=s.nextInt();
		int lengthofn=0;
		int temp=n;
		while(temp>0) {
			temp=temp/10;
			lengthofn++;
		}
		StringBuilder first=new StringBuilder();
		int noofnine=(n/9);
		
		for(int i=0;i<noofnine;i++) {
			first.append(9);
		}
		StringBuilder second=new StringBuilder ();
		int remaining=n;
		if(lengthofn==1) {
			second.append(9);
		}
		else{
			while(remaining>=9) {
			remaining-=9;
			second.append(9);
			}
		}
		second.append(9);
		for(int i=0;i<noofnine;i++) {
			second.append(0);
		}
		if(remaining==0){
		    first.append(1);
		    second.append(9);
		}else{
		    first.append(10-remaining<5?remaining:10-remaining);
		    second.append(10-remaining>5?remaining:10-remaining);
		}
		out.println(first);
		out.println(second);
		//out.println(second.length());
		out.close();
	}
	
	
	public static PrintWriter out;
	
	public static class FastReader {
		BufferedReader br;
		StringTokenizer st;
		//it reads the data about the specified point and divide the data about it ,it is quite fast
		//than using direct 

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

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

		int nextInt() {
			return Integer.parseInt(next());//converts string to integer
		}

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (Exception e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}