#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
clock_t start = clock();
inline long long int max(long long int a, long long int b, long long int c) {
  return max(max(a, b), c);
}
inline long long int min(long long int a, long long int b, long long int c) {
  return min(min(a, b), c);
}
inline long long int max(long long int a, long long int b) {
  return (a > b) ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return (a < b) ? a : b;
}
inline long long int mul(long long int x, long long int y, long long int mod_) {
  return ((x % mod_) * 1LL * (y % mod_)) % mod_;
}
long long int power(long long int a, long long int n) {
  long long int p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long int powm(long long int a, long long int n, long long int mod_) {
  long long int p = 1;
  while (n) {
    if (n % 2) {
      p = mul(p, a, mod_);
    }
    n >>= 1;
    a = mul(a, a, mod_);
  }
  return p % mod_;
}
long long int powi(long long int a, long long int mod_) {
  return powm(a, mod_ - 2, mod_);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long int mn, mx;
  long long int n, m, k, t, i, j, sum = 0, flag = 0, cnt = 0;
  long long int x = 0, y = 0, z, l = 0, r = 0, q;
  int TC = 1;
  while (TC--) {
    cin >> n >> k;
    x = (n - k) / 2;
    string ans;
    while ((long long int)ans.size() < n) {
      for (i = 0; i < x; ++i) ans.push_back('0');
      ans.push_back('1');
    }
    while ((long long int)ans.size() > n) ans.pop_back();
    cout << ans;
  };
  return 0;
}

### 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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
clock_t start = clock();
inline long long int max(long long int a, long long int b, long long int c) {
  return max(max(a, b), c);
}
inline long long int min(long long int a, long long int b, long long int c) {
  return min(min(a, b), c);
}
inline long long int max(long long int a, long long int b) {
  return (a > b) ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return (a < b) ? a : b;
}
inline long long int mul(long long int x, long long int y, long long int mod_) {
  return ((x % mod_) * 1LL * (y % mod_)) % mod_;
}
long long int power(long long int a, long long int n) {
  long long int p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long int powm(long long int a, long long int n, long long int mod_) {
  long long int p = 1;
  while (n) {
    if (n % 2) {
      p = mul(p, a, mod_);
    }
    n >>= 1;
    a = mul(a, a, mod_);
  }
  return p % mod_;
}
long long int powi(long long int a, long long int mod_) {
  return powm(a, mod_ - 2, mod_);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long int mn, mx;
  long long int n, m, k, t, i, j, sum = 0, flag = 0, cnt = 0;
  long long int x = 0, y = 0, z, l = 0, r = 0, q;
  int TC = 1;
  while (TC--) {
    cin >> n >> k;
    x = (n - k) / 2;
    string ans;
    while ((long long int)ans.size() < n) {
      for (i = 0; i < x; ++i) ans.push_back('0');
      ans.push_back('1');
    }
    while ((long long int)ans.size() > n) ans.pop_back();
    cout << ans;
  };
  return 0;
}

```