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;
bool ans[1024][1024];
void merge(int x1,int y1,int x2,int y2){
	if(x1==x2 || y1==y2) return;
	int x=(x1+x2)/2,y=(y1+y2)/2;
	ans[x][y]=0;
	merge(x1,y1,x-1,y-1);
	merge(x1,y+1,x-1,y2);
	merge(x+1,y1,x2,y-1);
	merge(x+1,y+1,x2,y2);
}
int main(){
	int n,m;
	cin>>n>>m;
	n=(1<<n)-1;m=(1<<m)-1;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=m;++j)
			ans[i][j]=1;
	merge(1,1,n,m);
	for(int i=1;i<=n;++i,puts(""))
		for(int j=1;j<=m;++j)
			cout<<int(ans[i][j]);
	return 0;
}