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 a[1025][1025],i,j,k,b[1025][1025],n,m;
bool flag;
int main()
{
	scanf("%d %d",&n,&m);
	if(n<m)
	{
		swap(n,m);
		flag=true;
	}
	a[1][1]=1;
	for(i=1;i<m;++i)
		for(j=0;j<(1<<i);++j)
			for(k=0;k<(1<<i);++k)
			{
				a[j+(1<<i)][k]=a[j][k+(1<<i)]=a[j][k];
				a[j+(1<<i)][k+(1<<i)]=a[j][k]^1;
			}
	for(i=(1<<m);i<(1<<n);++i)
		for(j=0;j<(1<<m);++j)
			a[i][j]=a[i-(1<<m)][j];
	for(i=(1<<n)-1;i>0;--i)
		for(j=(1<<m)-1;j>0;--j)
			a[i][j]^=a[i-1][j]^a[i][j-1]^a[i-1][j-1];
	if(flag)
	{
		for(i=1;i<(1<<n);++i)
			for(j=1;j<(1<<m);++j)
				b[j][i]=a[i][j];
		swap(n,m);
		for(i=1;i<(1<<n);++i)
			for(j=1;j<(1<<m);++j)
				a[i][j]=b[i][j];
	}
	for(i=1;i<(1<<n);++i)
	{
		for(j=1;j<(1<<m);++j)
			printf("%d",a[i][j]);
		printf("\n");
	}
}