#include <bits/stdc++.h>
using namespace std;
template <class n, class second>
ostream &operator<<(ostream &p, pair<n, second> x) {
  return p << "<" << x.first << ", " << x.second << ">";
}
template <class n>
auto operator<<(ostream &p, n y) ->
    typename enable_if<!is_same<n, string>::value,
                       decltype(y.begin(), p)>::type {
  int o = 0;
  p << "{";
  for (auto c : y) {
    if (o++) p << ", ";
    p << c;
  }
  return p << "}";
}
void dor() { cerr << '\n'; }
template <class n, class... second>
void dor(n p, second... y) {
  cerr << p << " ";
  dor(y...);
}
template <class n, class second>
void mini(n &p, second y) {
  if (p > y) p = y;
}
template <class n, class second>
void maxi(n &p, second y) {
  if (p < y) p = y;
}
bool correct(int n, int k, string second) {
  map<string, int> M1, M2;
  for (int i = 0; i < (int)(n - k + 1); ++i) {
    string s1 = second.substr(i, k);
    M1[s1]++;
  }
  for (int i = 0; i < (int)(n - k + 2); ++i) {
    string s1 = second.substr(i, k - 1);
    M2[s1]++;
  }
  int mi = 10;
  for (auto i : M1) mini(mi, i.second);
  int mi2 = 10;
  for (auto i : M2) mini(mi2, i.second);
  if (mi == 1 && mi2 != 1) return 1;
  return 0;
}
string br(int n, int k) {
  pair<int, string> best;
  best.first = -1e9;
  for (int i = 0; i < (int)(1 << n); ++i) {
    string second;
    for (int j = 0; j < (int)(n); ++j)
      if (i & (1 << j))
        second += "1";
      else
        second += "0";
    if (correct(n, k, second))
      mini(best, make_pair(-__builtin_popcount(i), second));
  }
  return best.second;
}
string sol(int n, int k) {
  int zer = (n - k) / 2;
  string second;
  while (second.size() < n) {
    for (int i = 0; i < (int)(zer); ++i) second += '0';
    second += '1';
  }
  while (second.size() > n) second.pop_back();
  return second;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  cout << sol(n, k) << '\n';
}

### 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;
template <class n, class second>
ostream &operator<<(ostream &p, pair<n, second> x) {
  return p << "<" << x.first << ", " << x.second << ">";
}
template <class n>
auto operator<<(ostream &p, n y) ->
    typename enable_if<!is_same<n, string>::value,
                       decltype(y.begin(), p)>::type {
  int o = 0;
  p << "{";
  for (auto c : y) {
    if (o++) p << ", ";
    p << c;
  }
  return p << "}";
}
void dor() { cerr << '\n'; }
template <class n, class... second>
void dor(n p, second... y) {
  cerr << p << " ";
  dor(y...);
}
template <class n, class second>
void mini(n &p, second y) {
  if (p > y) p = y;
}
template <class n, class second>
void maxi(n &p, second y) {
  if (p < y) p = y;
}
bool correct(int n, int k, string second) {
  map<string, int> M1, M2;
  for (int i = 0; i < (int)(n - k + 1); ++i) {
    string s1 = second.substr(i, k);
    M1[s1]++;
  }
  for (int i = 0; i < (int)(n - k + 2); ++i) {
    string s1 = second.substr(i, k - 1);
    M2[s1]++;
  }
  int mi = 10;
  for (auto i : M1) mini(mi, i.second);
  int mi2 = 10;
  for (auto i : M2) mini(mi2, i.second);
  if (mi == 1 && mi2 != 1) return 1;
  return 0;
}
string br(int n, int k) {
  pair<int, string> best;
  best.first = -1e9;
  for (int i = 0; i < (int)(1 << n); ++i) {
    string second;
    for (int j = 0; j < (int)(n); ++j)
      if (i & (1 << j))
        second += "1";
      else
        second += "0";
    if (correct(n, k, second))
      mini(best, make_pair(-__builtin_popcount(i), second));
  }
  return best.second;
}
string sol(int n, int k) {
  int zer = (n - k) / 2;
  string second;
  while (second.size() < n) {
    for (int i = 0; i < (int)(zer); ++i) second += '0';
    second += '1';
  }
  while (second.size() > n) second.pop_back();
  return second;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  cout << sol(n, k) << '\n';
}

```