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 n, m, Pow[20];
int a[1100][1100];

void build(int sx, int sy, int ex, int ey)
{
	if(sx == ex && sy == ey)
	{
		a[sx][sy] = 1;
		return;
	}
	int mix = (sx + ex) >> 1, miy = (sy + ey) >> 1;
	build(sx, sy, mix - 1, miy - 1);
	build(sx, miy + 1, mix - 1, ey);
	build(mix + 1, sy, ex, miy - 1);
	build(mix + 1, miy + 1, ex, ey);
	a[mix][miy] = 1;
}

int main()
{
	Pow[0] = 1; for(int i = 1; i < 20; ++i) Pow[i] = Pow[i - 1] * 2;
	scanf("%d %d", &n, &m); int k = max(n, m);
	build(1, 1, Pow[k] - 1, Pow[k] - 1);
	for(int i = 1; i <= Pow[n] - 1; ++i, puts(""))
		for(int j = 1; j <= Pow[m] - 1; ++j) printf("%d", a[i][j]);
	return 0;
}