Let s be some string consisting of symbols "0" or "1". Let's call a string t a substring of string s, if there exists such number 1 ≤ l ≤ |s| - |t| + 1 that t = s_l s_{l+1} … s_{l + |t| - 1}. Let's call a substring t of string s unique, if there exist only one such l. 

For example, let s = "1010111". A string t = "010" is an unique substring of s, because l = 2 is the only one suitable number. But, for example t = "10" isn't a unique substring of s, because l = 1 and l = 3 are suitable. And for example t ="00" at all isn't a substring of s, because there is no suitable l.

Today Vasya solved the following problem at the informatics lesson: given a string consisting of symbols "0" and "1", the task is to find the length of its minimal unique substring. He has written a solution to this problem and wants to test it. He is asking you to help him.

You are given 2 positive integers n and k, such that (n mod 2) = (k mod 2), where (x mod 2) is operation of taking remainder of x by dividing on 2. Find any string s consisting of n symbols "0" or "1", such that the length of its minimal unique substring is equal to k.

Input

The first line contains two integers n and k, separated by spaces (1 ≤ k ≤ n ≤ 100 000, (k mod 2) = (n mod 2)).

Output

Print a string s of length n, consisting of symbols "0" and "1". Minimal length of the unique substring of s should be equal to k. You can find any suitable string. It is guaranteed, that there exists at least one such string.

Examples

Input


4 4


Output


1111

Input


5 3


Output


01010

Input


7 3


Output


1011011

Note

In the first test, it's easy to see, that the only unique substring of string s = "1111" is all string s, which has length 4.

In the second test a string s = "01010" has minimal unique substring t ="101", which has length 3.

In the third test a string s = "1011011" has minimal unique substring t ="110", which has length 3.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author cplayer
 */
public class Main {
    public static void main (String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DTheMinimalUniqueSubstring solver = new DTheMinimalUniqueSubstring();
        solver.solve(1, in, out);
        out.close();
    }

    static class DTheMinimalUniqueSubstring {
        public void solve (int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt(), k = in.nextInt();
            // if (k == 1) {
            //     StringBuilder sb = new StringBuilder();
            //     sb.append(0);
            //     for (int i = 0; i < n - 1; ++i) {
            //         sb.append(1);
            //     }
            //     out.println(sb.toString());
            // } else {
            int l = (n - k + 2) / 2;
            StringBuilder sb = new StringBuilder();
            while (sb.length() < n) {
                for (int i = 0; i < l - 1 && sb.length() < n; ++i) {
                    sb.append(0);
                }
                if (sb.length() < n) {
                    sb.append(1);
                }
            }
            out.println(sb.toString());
            // }
        }

    }

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

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

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

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

    }
}

