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.
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
using pll = pair<ll, ll>;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using vvvll = vector<vector<vector<ll>>>;
using vii = vector<int>;
using vvii = vector<vector<int>>;
int ctoi(const char c) {
  if ('0' <= c && c <= '9') return (c - '0');
  return -1;
}
ll gcd(ll a, ll b) { return (b == 0 ? a : gcd(b, a % b)); }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
constexpr ll MOD = 1000000007;
constexpr ll INF = 1000000011;
constexpr ll MOD2 = 998244353;
constexpr ll LINF = 1001002003004005006ll;
constexpr ld EPS = 10e-10;
template <class T, class U>
inline bool chmax(T &lhs, const U &rhs) {
  if (lhs < rhs) {
    lhs = rhs;
    return 1;
  }
  return 0;
}
template <class T, class U>
inline bool chmin(T &lhs, const U &rhs) {
  if (lhs > rhs) {
    lhs = rhs;
    return 1;
  }
  return 0;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &&x : v) is >> x;
  return is;
}
template <typename T, typename U>
istream &operator>>(istream &is, pair<T, U> &p) {
  is >> p.first;
  is >> p.second;
  return is;
}
template <typename T, typename U>
ostream &operator>>(ostream &os, const pair<T, U> &p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  for (auto i = begin(v); i != end(v); ++i) {
    if (i != begin(v)) os << ' ';
    os << *i;
  }
  return os;
}
int main() {
  int N, M;
  cin >> N >> M;
  if (N == M) {
    for (ll i = 0; i < (N); ++i) {
      cout << 1;
    }
    cout << '\n';
    return 0;
  }
  if (M == 1) {
    cout << 0;
    for (ll i = 0; i < (N - 1); ++i) {
      cout << 1;
    }
    cout << '\n';
    return 0;
  }
  if (M == 2) {
    cout << "00";
    for (ll i = 0; i < (N - 2); ++i) {
      cout << 1;
    }
    cout << '\n';
    return 0;
  }
  if (N % 2 == 0) {
    int K = N / 2 - (M - 2) / 2;
    for (ll i = 0; i < (N); ++i) {
      if (i % K == 0) {
        cout << 0;
      } else {
        cout << 1;
      }
    }
    cout << '\n';
  } else {
    int K = N / 2 - (M - 2) / 2;
    for (ll i = 0; i < (N); ++i) {
      if (i % K == 0) {
        cout << 0;
      } else {
        cout << 1;
      }
    }
    cout << '\n';
  }
}
