#include <bits/stdc++.h>
using namespace std;
using Int = long long;
constexpr static int mod = 1e9 + 7;
constexpr static int inf = (1 << 30) - 1;
constexpr static Int infll = (1LL << 61) - 1;
int Competitive_Programming =
    (ios_base::sync_with_stdio(false), cin.tie(nullptr),
     cout << fixed << setprecision(15), 0);
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
bool check(string s, int N, int K) {
  if (N != s.size()) return false;
  map<string, int> mp;
  for (int j = 0; j < N; j++) {
    for (int l = j; l < N; l++) {
      auto q = s.substr(j, l - j + 1);
      mp[q]++;
    }
  }
  int mx = inf;
  for (auto &q : mp) {
    if (q.second == 1) mx = min(mx, (int)q.first.size());
  }
  if (mx % 2 != N % 2) return false;
  return mx == K;
}
int ex() {
  int N;
  cin >> N;
  vector<string> ts;
  for (int i = 0; i < (1 << N); i++) {
    string s = "";
    int k = i, j = N;
    while (j > 0) {
      j--;
      s.push_back((char)('0' + k % 2));
      k /= 2;
    }
    ts.push_back(s);
  }
  for (int i = 0; i < ts.size(); i++) {
    map<string, int> mp;
    for (int j = 0; j < N; j++) {
      for (int l = j; l < N; l++) {
        auto q = ts[i].substr(j, l - j + 1);
        mp[q]++;
      }
    }
    int mx = inf;
    for (auto &q : mp) {
      if (q.second == 1) mx = min(mx, (int)q.first.size());
    }
    if (mx % 2 != N % 2) continue;
    cout << ts[i] << ':';
    cout << mx << "\n";
  }
}
int main() {
  int N, K;
  cin >> N >> K;
  if (N == K) {
    for (int i = 0; i < N; i++) {
      cout << 1;
    }
    cout << '\n';
    return 0;
  }
  if (N % 2 == 1 and K == 1) {
    cout << 1;
    for (int i = 1; i < N; i++) {
      cout << 0;
    }
    cout << '\n';
    return 0;
  }
  string ans = "";
  if (N % 2 == 0) {
    int L = (N - K) / 2;
    while (ans.size() < N) {
      ans.push_back('1');
      for (int i = 0; i < L and ans.size() < N; i++) {
        ans.push_back('0');
      }
    }
  } else {
    int L = (N - K) / 2;
    int R = N % (L + 1);
    for (int i = 0; i < R; i++) {
      ans.push_back('0');
    }
    while (ans.size() < N) {
      ans.push_back('1');
      for (int i = 0; i < L and ans.size() < N; i++) {
        ans.push_back('0');
      }
    }
  }
  cout << ans << "\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;
using Int = long long;
constexpr static int mod = 1e9 + 7;
constexpr static int inf = (1 << 30) - 1;
constexpr static Int infll = (1LL << 61) - 1;
int Competitive_Programming =
    (ios_base::sync_with_stdio(false), cin.tie(nullptr),
     cout << fixed << setprecision(15), 0);
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
bool check(string s, int N, int K) {
  if (N != s.size()) return false;
  map<string, int> mp;
  for (int j = 0; j < N; j++) {
    for (int l = j; l < N; l++) {
      auto q = s.substr(j, l - j + 1);
      mp[q]++;
    }
  }
  int mx = inf;
  for (auto &q : mp) {
    if (q.second == 1) mx = min(mx, (int)q.first.size());
  }
  if (mx % 2 != N % 2) return false;
  return mx == K;
}
int ex() {
  int N;
  cin >> N;
  vector<string> ts;
  for (int i = 0; i < (1 << N); i++) {
    string s = "";
    int k = i, j = N;
    while (j > 0) {
      j--;
      s.push_back((char)('0' + k % 2));
      k /= 2;
    }
    ts.push_back(s);
  }
  for (int i = 0; i < ts.size(); i++) {
    map<string, int> mp;
    for (int j = 0; j < N; j++) {
      for (int l = j; l < N; l++) {
        auto q = ts[i].substr(j, l - j + 1);
        mp[q]++;
      }
    }
    int mx = inf;
    for (auto &q : mp) {
      if (q.second == 1) mx = min(mx, (int)q.first.size());
    }
    if (mx % 2 != N % 2) continue;
    cout << ts[i] << ':';
    cout << mx << "\n";
  }
}
int main() {
  int N, K;
  cin >> N >> K;
  if (N == K) {
    for (int i = 0; i < N; i++) {
      cout << 1;
    }
    cout << '\n';
    return 0;
  }
  if (N % 2 == 1 and K == 1) {
    cout << 1;
    for (int i = 1; i < N; i++) {
      cout << 0;
    }
    cout << '\n';
    return 0;
  }
  string ans = "";
  if (N % 2 == 0) {
    int L = (N - K) / 2;
    while (ans.size() < N) {
      ans.push_back('1');
      for (int i = 0; i < L and ans.size() < N; i++) {
        ans.push_back('0');
      }
    }
  } else {
    int L = (N - K) / 2;
    int R = N % (L + 1);
    for (int i = 0; i < R; i++) {
      ans.push_back('0');
    }
    while (ans.size() < N) {
      ans.push_back('1');
      for (int i = 0; i < L and ans.size() < N; i++) {
        ans.push_back('0');
      }
    }
  }
  cout << ans << "\n";
}

```