#include <bits/stdc++.h>
using namespace std;
template <class T, class T2>
inline int chkmax(T &x, const T2 &y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class T2>
inline int chkmin(T &x, const T2 &y) {
  return x > y ? x = y, 1 : 0;
}
const int MAXN = (1 << 20);
int n, k;
void read() { cin >> n >> k; }
void solve() {
  if (n == k) {
    for (int i = 1; i <= n; i++) cout << "1";
    cout << '\n';
    return;
  }
  if (k == 1) {
    for (int i = 0; i < n - 1; i++) cout << 1;
    cout << 0;
    cout << '\n';
    return;
  }
  if (k + 2 == n) {
    int bit = 1;
    for (int i = 1; i <= n; i++) {
      cout << bit;
      bit ^= 1;
    }
    cout << '\n';
    return;
  }
  if (k == 2) {
    cout << 0;
    for (int v = 0; v < n - 2; v++) cout << 1;
    cout << 0;
    cout << '\n';
    return;
  }
  for (int pref = 0; pref <= k - 2; pref++)
    for (int p = 2; p * (pref + 1) <= n; p++) {
      int l = k - 2 - pref;
      if (p != 2) {
        if (l % (p - 2) != 0) continue;
        l /= (p - 2);
        if (l == 1) continue;
        if (l * p + pref != n) continue;
        string base = "";
        for (int i = 0; i < l - 1; i++) base.push_back('1');
        base.push_back('0');
        for (int i = 0; i < p; i++) cout << base;
        for (int i = 0; i < pref; i++) cout << base[i];
        cout << '\n';
        return;
      } else if (l == 0) {
        if ((n - pref) % p != 0) continue;
        l = (n - pref) / p;
        if (l == 1) continue;
        if (l * p + pref != n) continue;
        string base = "";
        for (int i = 0; i < l - 1; i++) base.push_back('1');
        base.push_back('0');
        for (int i = 0; i < p; i++) cout << base;
        for (int i = 0; i < pref; i++) cout << base[i];
        cout << '\n';
        return;
      }
    }
  assert(false);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <class T, class T2>
inline int chkmax(T &x, const T2 &y) {
  return x < y ? x = y, 1 : 0;
}
template <class T, class T2>
inline int chkmin(T &x, const T2 &y) {
  return x > y ? x = y, 1 : 0;
}
const int MAXN = (1 << 20);
int n, k;
void read() { cin >> n >> k; }
void solve() {
  if (n == k) {
    for (int i = 1; i <= n; i++) cout << "1";
    cout << '\n';
    return;
  }
  if (k == 1) {
    for (int i = 0; i < n - 1; i++) cout << 1;
    cout << 0;
    cout << '\n';
    return;
  }
  if (k + 2 == n) {
    int bit = 1;
    for (int i = 1; i <= n; i++) {
      cout << bit;
      bit ^= 1;
    }
    cout << '\n';
    return;
  }
  if (k == 2) {
    cout << 0;
    for (int v = 0; v < n - 2; v++) cout << 1;
    cout << 0;
    cout << '\n';
    return;
  }
  for (int pref = 0; pref <= k - 2; pref++)
    for (int p = 2; p * (pref + 1) <= n; p++) {
      int l = k - 2 - pref;
      if (p != 2) {
        if (l % (p - 2) != 0) continue;
        l /= (p - 2);
        if (l == 1) continue;
        if (l * p + pref != n) continue;
        string base = "";
        for (int i = 0; i < l - 1; i++) base.push_back('1');
        base.push_back('0');
        for (int i = 0; i < p; i++) cout << base;
        for (int i = 0; i < pref; i++) cout << base[i];
        cout << '\n';
        return;
      } else if (l == 0) {
        if ((n - pref) % p != 0) continue;
        l = (n - pref) / p;
        if (l == 1) continue;
        if (l * p + pref != n) continue;
        string base = "";
        for (int i = 0; i < l - 1; i++) base.push_back('1');
        base.push_back('0');
        for (int i = 0; i < p; i++) cout << base;
        for (int i = 0; i < pref; i++) cout << base[i];
        cout << '\n';
        return;
      }
    }
  assert(false);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}

```