#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int N = 1e7 + 100;
string fun(string &s) {
  int n = s.length();
  for (int len = 1; len <= s.length(); len++) {
    map<string, int> mp;
    for (int j = 0; j <= n - len; j++) {
      string tmp;
      for (int i = j; i < j + len; i++) {
        tmp += s[i];
      }
      mp[tmp]++;
    }
    for (auto x : mp) {
      if (x.second == 1) {
        return x.first;
      }
    }
  }
}
bool is_palindrome(string s) {
  int n = s.length();
  for (int i = 0; i < n; i++) {
    if (s[i] != s[n - i - 1]) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  string ans;
  int a = (n - k) / 2;
  a++;
  for (int i = 1; i <= n; i++) {
    if (i % (a) == 0) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  return 0;
  int mi = 1e9;
  string tt;
  for (int i = 0; i < k; i++) {
    tt += '0';
  }
  for (int i = 0; i < (1 << n); i++) {
    string s;
    for (int j = 0; j < n; j++) {
      if ((1 << j) & i) {
        s += '1';
      } else {
        s += '0';
      }
    }
    if (!is_palindrome(s)) {
      continue;
    }
    if (fun(s).length() == k && s.substr((n - k) / 2, k) == fun(s)) {
      cout << s << "\n";
      cout << fun(s) << "\n";
      cout << "\n";
    }
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int N = 1e7 + 100;
string fun(string &s) {
  int n = s.length();
  for (int len = 1; len <= s.length(); len++) {
    map<string, int> mp;
    for (int j = 0; j <= n - len; j++) {
      string tmp;
      for (int i = j; i < j + len; i++) {
        tmp += s[i];
      }
      mp[tmp]++;
    }
    for (auto x : mp) {
      if (x.second == 1) {
        return x.first;
      }
    }
  }
}
bool is_palindrome(string s) {
  int n = s.length();
  for (int i = 0; i < n; i++) {
    if (s[i] != s[n - i - 1]) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  string ans;
  int a = (n - k) / 2;
  a++;
  for (int i = 1; i <= n; i++) {
    if (i % (a) == 0) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  return 0;
  int mi = 1e9;
  string tt;
  for (int i = 0; i < k; i++) {
    tt += '0';
  }
  for (int i = 0; i < (1 << n); i++) {
    string s;
    for (int j = 0; j < n; j++) {
      if ((1 << j) & i) {
        s += '1';
      } else {
        s += '0';
      }
    }
    if (!is_palindrome(s)) {
      continue;
    }
    if (fun(s).length() == k && s.substr((n - k) / 2, k) == fun(s)) {
      cout << s << "\n";
      cout << fun(s) << "\n";
      cout << "\n";
    }
  }
}

```