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 N,M,ans[1025][1025];
inline void solve(int lx,int rx,int ly,int ry)
{
	if(lx==rx) return;
	int midx=(lx+rx)>>1,midy=(ly+ry)>>1;
	ans[midx][midy]=0;
	solve(lx,midx-1,ly,midy-1);
	solve(lx,midx-1,midy+1,ry);
	solve(midx+1,rx,ly,midy-1);
	solve(midx+1,rx,midy+1,ry);
}
int main()
{
	cin>>N>>M;
	for(int i=1;i<(1<<N);i++) for(int j=1;j<(1<<M);j++) ans[i][j]=1;
	int T=max(N,M);solve(1,(1<<T)-1,1,(1<<T)-1);
	for(int i=1;i<(1<<N);i++){for(int j=1;j<(1<<M);j++) putchar(ans[i][j]+'0');putchar('\n');}
	return 0;
}