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;

const int N = 1024;

int ans[N][N];

int main() {
  for (int n = 1; n * 2 <= N; n *= 2) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        ans[i+n][j] = ans[i][j];
        ans[i][j+n] = ans[i][j];
        ans[i+n][j+n] = !ans[i][j];
      }
    }
  }
  for (int i = N-1; i > 0; --i) {
    for (int j = N-1; j > 0; --j) {
      ans[i][j] ^= ans[i-1][j-1] ^ ans[i-1][j] ^ ans[i][j-1];
    }
  }
  int n, m;
  scanf("%d %d", &n, &m);
  n = 1 << n;
  m = 1 << m;
  for (int i = 1; i < n; ++i) {
    for (int j = 1; j < m; ++j) {
      printf("%d", ans[i][j]);
    }
    printf("\n");
  }
  return 0;
}
