#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
const vector<int> dy = {-1, 0, 1, 0}, dx = {0, -1, 0, 1};
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
void debag(const vector<T> &a) {
  cerr << "debag :";
  for (auto v : a) cerr << v << " ";
  cerr << "\n";
}
template <class T>
void print(const vector<T> &a) {
  for (auto v : a) cout << v << " ";
  cout << "\n";
}
void solve() {
  ll n, k;
  cin >> n >> k;
  if (n == k) {
    cout << string(n, '1') << "\n";
    return;
  }
  if (k == 1) {
    cout << '1' << string(n - 1, '0') << "\n";
    return;
  }
  string res = {};
  if (k - 2 <= (n - k) / 2) {
    n -= k;
    k -= 2;
    res.push_back('0');
    for (int i = 0; i < k; i++) res.push_back('1');
    res.push_back('0');
    cout << string(n / 2, '1') << res << string(n / 2, '1') << "\n";
  } else if (0) {
    int v = (n - k) / 2;
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    int cnt = 0;
    string f = {};
    while (1) {
      if (int(f.size()) + int(res.size()) >= n) break;
      if (cnt == 0) {
        f.push_back('0');
        res.push_back('0');
      } else {
        f.push_back('1');
        res.push_back('1');
      }
      cnt++;
      if (cnt > v) cnt = 0;
    }
    if (int(f.size()) + int(res.size()) > n) res.pop_back();
    cout << f << res << "\n";
    f += res;
    res = f;
  } else {
    int v = (n - k) / 2;
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    int cnt = 0;
    string f = "0";
    while (1) {
      if (int(f.size()) + int(res.size()) >= n) break;
      if (cnt == 0) {
        res.push_back('0');
      } else {
        res.push_back('1');
      }
      cnt++;
      if (cnt > v) cnt = 0;
    }
    if (int(f.size()) + int(res.size()) > n) res.pop_back();
    cout << f << res << "\n";
    f += res;
    res = f;
  }
}
int main() {
  int t = 1;
  while (t--) solve();
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
const vector<int> dy = {-1, 0, 1, 0}, dx = {0, -1, 0, 1};
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
void debag(const vector<T> &a) {
  cerr << "debag :";
  for (auto v : a) cerr << v << " ";
  cerr << "\n";
}
template <class T>
void print(const vector<T> &a) {
  for (auto v : a) cout << v << " ";
  cout << "\n";
}
void solve() {
  ll n, k;
  cin >> n >> k;
  if (n == k) {
    cout << string(n, '1') << "\n";
    return;
  }
  if (k == 1) {
    cout << '1' << string(n - 1, '0') << "\n";
    return;
  }
  string res = {};
  if (k - 2 <= (n - k) / 2) {
    n -= k;
    k -= 2;
    res.push_back('0');
    for (int i = 0; i < k; i++) res.push_back('1');
    res.push_back('0');
    cout << string(n / 2, '1') << res << string(n / 2, '1') << "\n";
  } else if (0) {
    int v = (n - k) / 2;
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    int cnt = 0;
    string f = {};
    while (1) {
      if (int(f.size()) + int(res.size()) >= n) break;
      if (cnt == 0) {
        f.push_back('0');
        res.push_back('0');
      } else {
        f.push_back('1');
        res.push_back('1');
      }
      cnt++;
      if (cnt > v) cnt = 0;
    }
    if (int(f.size()) + int(res.size()) > n) res.pop_back();
    cout << f << res << "\n";
    f += res;
    res = f;
  } else {
    int v = (n - k) / 2;
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    res.push_back('0');
    res += string(v, '1');
    int cnt = 0;
    string f = "0";
    while (1) {
      if (int(f.size()) + int(res.size()) >= n) break;
      if (cnt == 0) {
        res.push_back('0');
      } else {
        res.push_back('1');
      }
      cnt++;
      if (cnt > v) cnt = 0;
    }
    if (int(f.size()) + int(res.size()) > n) res.pop_back();
    cout << f << res << "\n";
    f += res;
    res = f;
  }
}
int main() {
  int t = 1;
  while (t--) solve();
}

```