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>

const int N=2005;

int n,m,a[N][N];

void solve(int x)
{
	if (x==1) {a[1][1]=1;return;}
	solve(x-1);
	int n=1<<(x-1);
	for (int i=1;i<n;i++)
		for (int j=1;j<n;j++)
			a[i+n][j]=a[i][j+n]=a[i+n][j+n]=a[i][j];
	a[n][n]=1;
}

int main()
{
	scanf("%d%d",&n,&m);
	solve(std::max(n,m));
	for (int i=1;i<(1<<n);i++)
	{
		for (int j=1;j<(1<<m);j++) putchar(a[i][j]+'0');
		puts("");
	}
	return 0;
}
