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
n, m = map(int, input().split())
ans = [0, 1]
for i in range(1, min(n, m)):
    k = 1 << i
    x = (1 << k) - 1
    slide = [a << k for a in ans]
    new_ans = [s | a for s, a in zip(slide, ans)]
    new_ans.extend(s | (a ^ x) for s, a in zip(slide, ans))
    ans = new_ans
if n > m:
    ans *= 1 << (n - m)
elif n < m:
    for i in range(n, m):
        k = 1 << i
        ans = [(a << k) | a for a in ans]

ans = [a0 ^ a1 for a0, a1 in zip(ans, ans[1:])]
ans = [a ^ (a >> 1) for a in ans]

print('\n'.join(map(('{:0' + str(2 ** m - 1) + 'b}').format, ans)))
