#include <bits/stdc++.h>
using namespace std;
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(long long &x, long long y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(long long &x, long long y) { return (y > x) ? x = y, 1 : 0; }
const int N = 2000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
namespace GEO {
struct Point {
  long long x, y;
  int id;
  Point() {}
  Point(long long x, long long y) : x(x), y(y) {}
  long double len() { return sqrt(x * x + y * y); }
  long long len2() { return x * x + y * y; }
  void read() { cin >> x >> y; }
};
bool operator==(Point p1, Point p2) { return p1.x == p2.x && p1.y == p2.y; }
bool operator!=(Point p1, Point p2) { return p1.x != p2.x || p1.y != p2.y; }
Point operator+(Point p1, Point p2) { return Point(p1.x + p2.x, p1.y + p2.y); }
Point operator-(Point p1, Point p2) { return Point(p1.x - p2.x, p1.y - p2.y); }
long long operator*(Point p1, Point p2) { return p1.x * p2.x + p1.y * p2.y; }
long long operator%(Point p1, Point p2) { return p1.x * p2.y - p1.y * p2.x; }
long double dist(Point p1, Point p2) { return (p1 - p2).len(); }
long long dist2(Point p1, Point p2) { return (p1 - p2).len2(); }
}  // namespace GEO
using namespace GEO;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int n, k;
  cin >> n >> k;
  int c = (n - k) / 2 + 1, p = 0;
  for (int i = 0; i < n; i++) {
    p = (p + 1) % c;
    if (p == 0) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  cout << "\n";
  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;
inline bool setmin(int &x, int y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(int &x, int y) { return (y > x) ? x = y, 1 : 0; }
inline bool setmin(long long &x, long long y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(long long &x, long long y) { return (y > x) ? x = y, 1 : 0; }
const int N = 2000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int P1 = 31;
const int P2 = 57;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const int MOD2 = 998244353;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
namespace GEO {
struct Point {
  long long x, y;
  int id;
  Point() {}
  Point(long long x, long long y) : x(x), y(y) {}
  long double len() { return sqrt(x * x + y * y); }
  long long len2() { return x * x + y * y; }
  void read() { cin >> x >> y; }
};
bool operator==(Point p1, Point p2) { return p1.x == p2.x && p1.y == p2.y; }
bool operator!=(Point p1, Point p2) { return p1.x != p2.x || p1.y != p2.y; }
Point operator+(Point p1, Point p2) { return Point(p1.x + p2.x, p1.y + p2.y); }
Point operator-(Point p1, Point p2) { return Point(p1.x - p2.x, p1.y - p2.y); }
long long operator*(Point p1, Point p2) { return p1.x * p2.x + p1.y * p2.y; }
long long operator%(Point p1, Point p2) { return p1.x * p2.y - p1.y * p2.x; }
long double dist(Point p1, Point p2) { return (p1 - p2).len(); }
long long dist2(Point p1, Point p2) { return (p1 - p2).len2(); }
}  // namespace GEO
using namespace GEO;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  int n, k;
  cin >> n >> k;
  int c = (n - k) / 2 + 1, p = 0;
  for (int i = 0; i < n; i++) {
    p = (p + 1) % c;
    if (p == 0) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
  cout << "\n";
  return 0;
}

```