#include <bits/stdc++.h>
using namespace std;
void gen(int n, int k, string r) {
  vector<string> dfr;
  for (int i = 0; i < (1 << n); i++) {
    string s = "";
    for (int j = 0; j < n; j++) {
      if (i & (1 << j))
        s += '1';
      else
        s += '0';
    }
    int mk = 0;
    for (int j = 1; j <= k && !mk; j++) {
      map<string, int> df;
      for (int p = 0; p <= n - j; p++) {
        string temp = s.substr(p, j);
        df[temp]++;
      }
      for (auto p = df.begin(); p != df.end(); p++) {
        if (p->second == 1) {
          if (j == k) {
            dfr.push_back(s);
          } else {
            mk = 1;
            break;
          }
        }
      }
    }
  }
  sort(dfr.begin(), dfr.end());
  int pres = 0;
  for (int i = 0; i < dfr.size(); i++) {
    if (dfr[i] == r) pres = 1;
  }
  if (pres)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  int x = (n - k) / 2;
  int ctr = 2;
  string s = "01";
  if (k == 1 && n == 1) {
    cout << 1 << '\n';
    return 0;
  }
  if (k == n) {
    for (int i = 0; i < n; i++) cout << 1;
    cout << '\n';
    return 0;
  }
  if (k == 2) {
    string ans = "11";
    while (ans.length() < n) {
      ans.push_back('0');
    }
    reverse(ans.begin(), ans.end());
    cout << ans << '\n';
    return 0;
  }
  if (k == 1) {
    string ans = "1";
    while (ans.length() < n) {
      ans.push_back('0');
    }
    reverse(ans.begin(), ans.end());
    cout << ans << '\n';
    return 0;
  }
  while (ctr < n) {
    for (int i = 0; i < x && ctr < n; i++) s.push_back('0'), ctr++;
    if (ctr < n) s.push_back('1'), ctr++;
  }
  reverse(s.begin(), s.end());
  cout << s << '\n';
}

### Prompt

Your task is to create a cpp solution to the following problem:

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;
void gen(int n, int k, string r) {
  vector<string> dfr;
  for (int i = 0; i < (1 << n); i++) {
    string s = "";
    for (int j = 0; j < n; j++) {
      if (i & (1 << j))
        s += '1';
      else
        s += '0';
    }
    int mk = 0;
    for (int j = 1; j <= k && !mk; j++) {
      map<string, int> df;
      for (int p = 0; p <= n - j; p++) {
        string temp = s.substr(p, j);
        df[temp]++;
      }
      for (auto p = df.begin(); p != df.end(); p++) {
        if (p->second == 1) {
          if (j == k) {
            dfr.push_back(s);
          } else {
            mk = 1;
            break;
          }
        }
      }
    }
  }
  sort(dfr.begin(), dfr.end());
  int pres = 0;
  for (int i = 0; i < dfr.size(); i++) {
    if (dfr[i] == r) pres = 1;
  }
  if (pres)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  int x = (n - k) / 2;
  int ctr = 2;
  string s = "01";
  if (k == 1 && n == 1) {
    cout << 1 << '\n';
    return 0;
  }
  if (k == n) {
    for (int i = 0; i < n; i++) cout << 1;
    cout << '\n';
    return 0;
  }
  if (k == 2) {
    string ans = "11";
    while (ans.length() < n) {
      ans.push_back('0');
    }
    reverse(ans.begin(), ans.end());
    cout << ans << '\n';
    return 0;
  }
  if (k == 1) {
    string ans = "1";
    while (ans.length() < n) {
      ans.push_back('0');
    }
    reverse(ans.begin(), ans.end());
    cout << ans << '\n';
    return 0;
  }
  while (ctr < n) {
    for (int i = 0; i < x && ctr < n; i++) s.push_back('0'), ctr++;
    if (ctr < n) s.push_back('1'), ctr++;
  }
  reverse(s.begin(), s.end());
  cout << s << '\n';
}

```