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.*;

public class B 
{
	
	static StringBuilder st ; 
	
	public static void main(String[] args) throws Exception
	{
		Scanner sc = new Scanner () ;
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		String first = "" ; 
		String second = "" ; 
		
		
		for(int i = 0 ; i < 200 ;i++ )
		{
			first+= "9";
			second+="9";
		}
		
		for(int i = 0 ; i < 199 ;i++ )
			first+= "0";
		
		first += "1";
		
		out.println(first+" "+second);
		
		out.flush();
		out.close();
	
	}
	
	static int sumDigits(int num)
	{
		int ans = 0 ;
		
		while(num > 0)
		{
			ans += num % 10  ;
			
			num /= 10;
		}
		
		return ans ;
		
	}
	
	static class Scanner 
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;

		String next() throws Exception 
		{
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());

			return st.nextToken();
		}

		int nextInt() throws Exception { return Integer.parseInt(next()); }
		
		long nextLong() throws Exception { return Long.parseLong(next()); }

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

		char nextChar() throws Exception { return next().charAt(0); }
		
		char [] nextCharArray() throws Exception{ return next().toCharArray() ; }
	}

}
