We have a grid with (2^N - 1) rows and (2^M-1) columns. You are asked to write 0 or 1 in each of these squares. Let a_{i,j} be the number written in the square at the i-th row from the top and the j-th column from the left.

For a quadruple of integers (i_1, i_2, j_1, j_2) such that 1\leq i_1 \leq i_2\leq 2^N-1, 1\leq j_1 \leq j_2\leq 2^M-1, let S(i_1, i_2, j_1, j_2) = \displaystyle \sum_{r=i_1}^{i_2}\sum_{c=j_1}^{j_2}a_{r,c}. Then, let the oddness of the grid be the number of quadruples (i_1, i_2, j_1, j_2) such that S(i_1, i_2, j_1, j_2) is odd.

Find a way to fill in the grid that maximizes its oddness.

Constraints

* N and M are integers between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:


N M


Output

Print numbers to write in the grid so that its oddness is maximized, in the following format:


a_{1,1}a_{1,2}\cdots a_{1,2^M-1}
a_{2,1}a_{2,2}\cdots a_{2,2^M-1}
\vdots
a_{2^N-1,1}a_{2^N-1,2}\cdots a_{2^N-1,2^M-1}


If there are multiple solutions, you can print any of them.

Example

Input

1 2


Output

111
#include <bits/stdc++.h>

using namespace std;

int main() {
   ios_base::sync_with_stdio(false);
   int N, M;
   cin >> N >> M;
   N = (1 << N) - 1;
   M = (1 << M) - 1;
   vector<vector<int>> ans(N, vector<int>(M));
   function<void(int, int, int, int)> go = [&](int xl, int xr, int yl, int yr) {
      int xm = (xl + xr) >> 1;
      int ym = (yl + yr) >> 1;
      ans[xm][ym] = 1;
      if (xl < xr && yl < yr) {
         go(xl, xm - 1, yl, ym - 1);
         go(xl, xm - 1, ym + 1, yr);
         go(xm + 1, xr, yl, ym - 1);
         go(xm + 1, xr, ym + 1, yr);
      }
   };
   go(0, N - 1, 0, M - 1);
   for (int i = 0; i < N; ++i) {
      for (int j = 0; j < M; ++j) {
         cout << ans[i][j];
      }
      cout << "\n";
   }
}
