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
//%std
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
	int out = 0, fh = 1;
	char jp = getchar();
	while ((jp > '9' || jp < '0') && jp != '-')
		jp = getchar();
	if (jp == '-')
		fh = -1, jp = getchar();
	while (jp >= '0' && jp <= '9')
		out = out * 10 + jp - '0', jp = getchar();
	return out * fh;
}
const int N = (1 << 10) + 10;
int a[11][N][N];
void construct(int n)
{
	if (n == 0)
		return;
	construct(n - 1);
	int len = 1 << (n - 1);
	for (int i = 1; i < len; ++i)
		for (int j = 1; j < len; ++j)
		{
			int t = a[n - 1][i][j];
			a[n][i][j] = t;
			a[n][i + len][j] = t;
			a[n][i][j + len] = t;
			a[n][i + len][j + len] = t;
		}		
	a[n][len][len] = 1;
}
int n, m, f = 0;
int main()
{
	n = read(), m = read();
	if (n < m)
		swap(n, m), f = 1;
	construct(n);
	if (!f)
	{
		for (int i = 1; i <= (1 << n) - 1; ++i, puts(""))
			for (int j = 1; j <= (1 << m) - 1; ++j)
				printf("%d", a[n][i][j]);
	}
	else
	{
		for (int j = 1; j <= (1 << m) - 1; ++j, puts(""))
			for (int i = 1; i <= (1 << n) - 1; ++i)
				printf("%d", a[n][i][j]);
	}
	return 0;
}
