#include <bits/stdc++.h>
const int oo = 0x7fffffff;
const long long OO = 0x7fffffffffffffff;
template <class T>
std::ostream& operator<<(std::ostream& O, const std::vector<T>& V) {
  for (const T& x : V) O << x << " ";
  O << "\n";
  return O;
}
template <class T1, class T2>
std::ostream& operator<<(std::ostream& O, const std::pair<T1, T2>& P) {
  return O << "(" << P.first << ", " << P.second << ")";
}
using namespace std;
string unique_s(string s) {
  map<string, int> ct;
  int n = s.size();
  for (int l = 1; l <= n; ++l) {
    ct.clear();
    for (int i = 0; (i + l) <= n; ++i) {
      auto t = s.substr(i, l);
      ct[t]++;
    }
    for (auto p : ct)
      if (p.second == 1) return p.first;
  }
  return string();
}
string solve(int N, int M) {
  string s;
  for (int i = 0; i < (1 << N); ++i) {
    s = (bitset<100>(i)).to_string().substr(100 - N);
    if (M == unique_s(s).size()) return s;
  }
  return string("-1");
}
string solve_2(int N, int M) {
  if (M == 1) return "1" + string(N - 1, '0');
  if (N == M) return string(N, '0');
  int l = M - 2;
  if ((3 * l + 2) <= N) {
    string z(M - 2, '0');
    return string(N - (2 * l + 2), '0') + "1" + z + "1" + z;
  }
  string z((N - M) / 2, '0');
  string s;
  while (s.size() < N) s += z + "1";
  return s.substr(0, N);
}
int main() {
  int n;
  cin >> n;
  int m;
  cin >> m;
  cout << solve_2(n, m) << "\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
const int oo = 0x7fffffff;
const long long OO = 0x7fffffffffffffff;
template <class T>
std::ostream& operator<<(std::ostream& O, const std::vector<T>& V) {
  for (const T& x : V) O << x << " ";
  O << "\n";
  return O;
}
template <class T1, class T2>
std::ostream& operator<<(std::ostream& O, const std::pair<T1, T2>& P) {
  return O << "(" << P.first << ", " << P.second << ")";
}
using namespace std;
string unique_s(string s) {
  map<string, int> ct;
  int n = s.size();
  for (int l = 1; l <= n; ++l) {
    ct.clear();
    for (int i = 0; (i + l) <= n; ++i) {
      auto t = s.substr(i, l);
      ct[t]++;
    }
    for (auto p : ct)
      if (p.second == 1) return p.first;
  }
  return string();
}
string solve(int N, int M) {
  string s;
  for (int i = 0; i < (1 << N); ++i) {
    s = (bitset<100>(i)).to_string().substr(100 - N);
    if (M == unique_s(s).size()) return s;
  }
  return string("-1");
}
string solve_2(int N, int M) {
  if (M == 1) return "1" + string(N - 1, '0');
  if (N == M) return string(N, '0');
  int l = M - 2;
  if ((3 * l + 2) <= N) {
    string z(M - 2, '0');
    return string(N - (2 * l + 2), '0') + "1" + z + "1" + z;
  }
  string z((N - M) / 2, '0');
  string s;
  while (s.size() < N) s += z + "1";
  return s.substr(0, N);
}
int main() {
  int n;
  cin >> n;
  int m;
  cin >> m;
  cout << solve_2(n, m) << "\n";
  return 0;
}

```