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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename T>
bool uax(T &a, const T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool uin(T &a, const T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
mt19937 rnd(1928);
int cnt_v(string &s, string t) {
  int ret = 0;
  for (int i = 0; i + (int)t.size() - 1 < s.size(); i++) {
    if (s.substr(i, t.size()) == t) {
      ret++;
    }
  }
  return ret;
}
int kog(string &s) {
  for (int i = 1; i <= s.size(); i++) {
    for (int j = 0; j + i - 1 < s.size(); j++) {
      if (cnt_v(s, s.substr(j, i)) == 1) return i;
    }
  }
}
void solv(int n, int k) {
  string s;
  s.resize(n);
  for (int i = 0; i < (1 << n); i++) {
    for (int j = 0; j < n; j++) {
      if ((1 << j) & i) {
        s[j] = '1';
      } else {
        s[j] = '0';
      }
    }
    if (kog(s) == k) {
      cout << s << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  if (k == 1) {
    cout << 1;
    for (int i = 0; i < n - 1; i++) {
      cout << 0;
    }
    exit(0);
  }
  if (n == k) {
    for (int i = 0; i < n; i++) {
      cout << 0;
    }
    exit(0);
  }
  int cnt = 0;
  bool boob = false;
  if (k * 3 - 4 <= n) {
    cnt = k - 2;
  } else {
    cnt = (n - k) / 2;
    boob = true;
  }
  string w = "";
  for (int j = 0; j < cnt; j++) {
    w.push_back('0');
  }
  w.push_back('1');
  int cnte = 0;
  for (int i = 0, j = 0; i < n; i++, j = (j + 1) % w.size()) {
    if (boob)
      cout << w[j];
    else {
      if (cnte >= 2)
        cout << 0;
      else
        cout << w[j];
    }
    if (w[j] == '1') cnte++;
  }
}
