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.*;
// import java.util.*;

// public class C2 {
//     public static void main(String[] args) throws IOException {
//         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
//         StringTokenizer st = new StringTokenizer(br.readLine());
//         int n = Integer.parseInt(st.nextToken());
//         int k = Integer.parseInt(st.nextToken());
//         if (n < (long) k * (k + 1) / 2 || k == 2 && n == 4 || k == 3 && n == 8) {
//             System.out.println("NO");
//             return;
//         }
//         System.out.println("YES");
//         if (k == 1) {
//             System.out.println(n);
//             return;
//         }
//         int[] aa = new int[1 + k];
//         n -= k * (k + 1) / 2;
//         int m = n / k, r = n % k;
//         for (int i = 1; i <= k; i++)
//             aa[i] = m + i + (i + r > k ? 1 : 0);
//         if (n == k - 1) {
//             aa[2]--;
//             aa[k]++;
//         }
//         StringBuilder sb = new StringBuilder();
//         for (int i = 1; i <= k; i++)
//             sb.append(aa[i] + " ");
//         System.out.println(sb);
//     }
// }package Round558;
// public class B1163{
// 	public static void main(String[] args) {
// 		Scanner sc=new Scanner(System.in);
// 		int n,i,ans=0,colour,max=0;
// 		n=sc.nextInt();
// 		int fr[]=new int[100001];
// 		int count[]=new int[100001];
// 		for(i=0;i<n;i++){
// 			colour=sc.nextInt();
// 			count[fr[colour]]--;
// 			fr[colour]++;
// 			count[fr[colour]]++;
// 			max=Math.max(max,fr[colour]);
// 			if((count[1]==i+1)
// 				||(count[i+1]==1)
// 				||((count[1]==1)&&((count[max]*max)==i))
// 				||((count[max]==1)&&(((count[max-1]+1)*(max-1))==i)))
// 				ans=i;
// 		}
// 		System.out.println(ans+1);
// 	}
// }
import java.util.Scanner;
import java.util.Arrays;
public class A{
	public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), k = in.nextInt();
        System.out.println(solve(n, k));
    }

    static String solve(int n, int k) {
        char[] s = new char[n];
        int d = (n - k) / 2;
        for (int i = 0, j = 0; i < n; i++) {
            if (i < d) s[i] = '1';
            else if (i == d) s[i] = '0';
            else s[i] = s[j++];
        }
        return new String(s);
    }
}