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<iostream>
#include<bitset>
#include<algorithm>
using namespace std;
bitset<1<<10>T[10];
int N,M;
bool tr=false;
bitset<1<<10>f[1<<10];
int A[1<<10][1<<10];
int main()
{
	cin>>N>>M;
	if(N<M)
	{
		tr=true;
		swap(N,M);
	}
	for(int i=0;i<N;i++)
	{
		for(int j=0;j<1<<N;j++)
		{
			T[i][j]=j>>i&1;
		}
	}
	for(int i=0;i<1<<M;i++)
	{
		for(int j=0;j<N;j++)
		{
			if(i>>j&1)f[i]^=T[j];
		}
	}
	for(int i=1;i<1<<N;i++)for(int j=1;j<1<<M;j++)
	{
		A[i][j]=f[j][i]^f[j-1][i]^f[j][i-1]^f[j-1][i-1];
	}
	if(tr)
	{
		for(int i=1;i<1<<M;i++)
		{
			for(int j=1;j<1<<N;j++)cout<<A[j][i];
			cout<<endl;
		}
	}
	else
	{
		for(int i=1;i<1<<N;i++)
		{
			for(int j=1;j<1<<M;j++)cout<<A[i][j];
			cout<<endl;
		}
	}
}
