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.
#include <bits/stdc++.h>
using namespace std;
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long int nxtl() {
  long long int x;
  cin >> x;
  return x;
}
void SieveOfEratosthenes(int n, vector<int> &primes) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int max(long long int a, long long int b) {
  if (a > b) return a;
  return b;
}
long long int power(long long int x, long long int y, long long int mod) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, mod);
  if (y % 2 == 0)
    return (temp * temp) % mod;
  else
    return (((x * temp) % mod) * temp) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n = nxt(), k = nxt();
  int a = (n - k) / 2;
  vector<int> temp;
  for (int i = 0; i < a; i++) temp.push_back(0);
  temp.push_back(1);
  int j = 0;
  while (n--) {
    cout << temp[j];
    j = (j + 1) % (a + 1);
  }
  cout << endl;
  return 0;
}
