#include <bits/stdc++.h>
using namespace std;
using lint = long long int;
using ll = long long int;
using lnt = long long int;
using Graph = vector<vector<int>>;
using Wgraph = vector<vector<pair<int, long long int>>>;
long long int INF = 1001001001001001LL;
int inf = 1000000007;
long long int MOD = 1000000007LL;
double PI = 3.1415926535897932;
long long int di[] = {-1, 0, 1, 0, -1, 1, 1, -1};
long long int dj[] = {0, 1, 0, -1, 1, 1, -1, -1};
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
inline void yes() { cout << "yes" << endl; }
inline void Yes() { cout << "Yes" << endl; }
inline void YES() { cout << "YES" << endl; }
inline void no() { cout << "no" << endl; }
inline void No() { cout << "No" << endl; }
inline void NO() { cout << "NO" << endl; }
inline void possible() { cout << "possible" << endl; }
inline void Possible() { cout << "Possible" << endl; }
inline void POSSIBLE() { cout << "POSSIBLE" << endl; }
inline void impossible() { cout << "impossible" << endl; }
inline void Impossible() { cout << "Impossible" << endl; }
inline void IMPOSSIBLE() { cout << "IMPOSSIBLE" << endl; }
int main() {
  int n, k;
  cin >> n >> k;
  lint len = (n - k) / 2;
  string ans = "";
  for (int i = 0; i < n; i++) {
    if (i % (len + 1) == len)
      ans += "1";
    else
      ans += "0";
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp to 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>
using namespace std;
using lint = long long int;
using ll = long long int;
using lnt = long long int;
using Graph = vector<vector<int>>;
using Wgraph = vector<vector<pair<int, long long int>>>;
long long int INF = 1001001001001001LL;
int inf = 1000000007;
long long int MOD = 1000000007LL;
double PI = 3.1415926535897932;
long long int di[] = {-1, 0, 1, 0, -1, 1, 1, -1};
long long int dj[] = {0, 1, 0, -1, 1, 1, -1, -1};
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
inline void yes() { cout << "yes" << endl; }
inline void Yes() { cout << "Yes" << endl; }
inline void YES() { cout << "YES" << endl; }
inline void no() { cout << "no" << endl; }
inline void No() { cout << "No" << endl; }
inline void NO() { cout << "NO" << endl; }
inline void possible() { cout << "possible" << endl; }
inline void Possible() { cout << "Possible" << endl; }
inline void POSSIBLE() { cout << "POSSIBLE" << endl; }
inline void impossible() { cout << "impossible" << endl; }
inline void Impossible() { cout << "Impossible" << endl; }
inline void IMPOSSIBLE() { cout << "IMPOSSIBLE" << endl; }
int main() {
  int n, k;
  cin >> n >> k;
  lint len = (n - k) / 2;
  string ans = "";
  for (int i = 0; i < n; i++) {
    if (i % (len + 1) == len)
      ans += "1";
    else
      ans += "0";
  }
  cout << ans << endl;
  return 0;
}

```