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>
#define SIZE 2000
using namespace std;
typedef pair <int,int> P;
typedef long long int ll;

int ans[SIZE][SIZE];

void solve(int n,int m)
{
	if(n<m)
	{
		solve(n,n);
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				ans[i][j]=ans[i][j%n];
			}
		}
	}
	else if(n>m)
	{
		solve(m,m);
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				ans[i][j]=ans[i%m][j];
			}
		}
	}
	else
	{
		if(n==1) ans[0][0]=0;
		else
		{
			solve(n/2,m/2);
			for(int i=0;i<n;i++)
			{
				for(int j=0;j<m;j++)
				{
					if(i<n/2||j<m/2) ans[i][j]=ans[i%(n/2)][j%(m/2)];
					else ans[i][j]=1-ans[i%(n/2)][j%(m/2)];
				}
			}
		}
	}
}
int main()
{
	int n,m;
	scanf("%d %d",&n,&m);
	solve(1<<n,1<<m);
	for(int i=0;i<(1<<n)-1;i++)
	{
		for(int j=0;j<(1<<m)-1;j++)
		{
			printf("%d",(ans[i+1][j+1]^ans[i][j]^ans[i+1][j]^ans[i][j+1]));
		}
		puts("");
	}
	return 0;
}
	