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 STDIO
#define MOD 1000000007
#define testbit(mask,i) (((mask)>>(i))&1)
#define setbit(mask,i) ((mask)|(1<<i))
#define delbit(mask,i) ((mask)^(1<<i))
#define lowbit(x) ((x)&(-x))
#define INF MOD
using namespace std;
int n,m;
int main(){
	#ifndef STDIO
		freopen("input.in","r",stdin);
		freopen("output.out","w",stdout);
	#endif
	ios::sync_with_stdio(0);
	cin.tie(NULL);
	cout.tie(NULL);
	cin>>n>>m;
	for(int i=1;i<(1<<n);i++){
		for(int j=1;j<(1<<m);j++){
			if(lowbit(i)==lowbit(j))
				cout<<1;
			else
				cout<<0;
		}
		cout<<endl;
	}
	return 0;
}
