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<cstdio>

int n, m;
bool ans[2000][2000];

void f(int x1, int x2, int y1, int y2) {
  if (x1 > x2 || y1 > y2) {
    return;
  }
  int midx = (x1 + x2) / 2;
  int midy = (y1 + y2) / 2;
  f(x1, midx - 1, y1, midy - 1);
  f(x1, midx - 1, midy + 1, y2);
  f(midx + 1, x2, y1, midy - 1);
  f(midx + 1, x2, midy + 1, y2);
  ans[midx][midy] = true;
}

int main() {
  f(1, 1023, 1, 1023);
  scanf("%d %d", &n, &m);
  for (int i = 1; i < (1 << n); i++) {
    for (int j = 1; j < (1 << m); j++) {
      printf("%d", ans[i][j]);
    }
    printf("\n");
  }
}
