#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
int gcd(int a, int b) {
  if (b) return gcd(b, a % b);
  return a;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, k;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n >> k;
  string s = "";
  if (n == k) {
    for (int i = 1; i <= n; i++) s += '1';
    return cout << s, 0;
  }
  if (k == 1) {
    cout << "1";
    for (int i = 1; i < n; i++) cout << "0";
    return 0;
  }
  if (k - 2 <= (n - k) / 2) {
    for (int i = 1; i <= (n - k) / 2; i++) s += '0';
    s += '1';
    for (int i = 1; i <= k - 2; i++) s += '0';
    s += '1';
    for (int i = 1; i <= (n - k) / 2; i++) s += '0';
    return cout << s, 0;
  } else {
    int f = 1 + (n - k) / 2;
    string l = "1";
    for (int i = 1; i < f; i++) l += '0';
    string ans = "";
    while ((int)((ans).size()) < n) ans += l;
    while ((int)((ans).size()) > n) ans.pop_back();
    return cout << ans, 0;
  }
  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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
std::pair<int, int> DR[] = {{-1, 0}, {0, 1},   {1, 0}, {0, -1},
                            {-1, 1}, {-1, -1}, {1, 1}, {1, -1}};
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
int gcd(int a, int b) {
  if (b) return gcd(b, a % b);
  return a;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int n, k;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n >> k;
  string s = "";
  if (n == k) {
    for (int i = 1; i <= n; i++) s += '1';
    return cout << s, 0;
  }
  if (k == 1) {
    cout << "1";
    for (int i = 1; i < n; i++) cout << "0";
    return 0;
  }
  if (k - 2 <= (n - k) / 2) {
    for (int i = 1; i <= (n - k) / 2; i++) s += '0';
    s += '1';
    for (int i = 1; i <= k - 2; i++) s += '0';
    s += '1';
    for (int i = 1; i <= (n - k) / 2; i++) s += '0';
    return cout << s, 0;
  } else {
    int f = 1 + (n - k) / 2;
    string l = "1";
    for (int i = 1; i < f; i++) l += '0';
    string ans = "";
    while ((int)((ans).size()) < n) ans += l;
    while ((int)((ans).size()) > n) ans.pop_back();
    return cout << ans, 0;
  }
  return 0;
}

```