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.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Unnatural {
    public static void main(String[] args) {

        FastReader in = new FastReader();
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        int i,j,a,b,c,n,k,l,m;
        long sum = 0;
        int[] brr = new int[1];
        int[] arr = new int[1];
        //mergeSort(arr, 0, 7);

        n = in.nextInt(); m = in.nextInt();

        for (i=1; i<=2230; i++)
            out.print(8);
        out.println();
        for (i=1; i<=2229; i++)
            out.print(1);
        out.println(2);
        out.flush();



    }



    static void  mergeSort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Same as (l+r)/2, but avoids overflow for
            // large l and h
            int m = l+(r-l)/2;

            // Sort first and second halves
            mergeSort(arr, l, m);
            mergeSort(arr, m+1, r);

            merge(arr, l, m, r);
        }
    }


    static void merge(int arr[], int l, int m, int r)
    {
        int i, j, k;
        int n1 = m - l + 1;
        int n2 =  r - m;

        /* create temp arrays */
        int L[] = new int[n1],  R[] = new int[n2];

        /* Copy data to temp arrays L[] and R[] */
        for (i = 0; i < n1; i++)
            L[i] = arr[l + i];
        for (j = 0; j < n2; j++)
            R[j] = arr[m + 1+ j];

        /* Merge the temp arrays back into arr[l..r]*/
        i = 0; // Initial index of first subarray
        j = 0; // Initial index of second subarray
        k = l; // Initial index of merged subarray
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

    /* Copy the remaining elements of L[], if there
       are any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

    /* Copy the remaining elements of R[], if there
       are any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }






    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

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

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

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

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

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

}