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<cstdio>
#include<cstring>
using namespace std;
#define N 1333
inline int read(){
	int x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
int a[N][N],n,m;
void Copy(int x,int y,int len){
	for(int i=1;i<=len;++i){
		for(int j=1;j<=len;++j){
			a[i+x][j+y]=a[i][j];
		}
	}
}
void dfs(int n){
	if(n==1)return (void)(a[1][1]=1);
	dfs(n>>1);
	Copy(0,(n>>1)+1,n>>1);
	Copy((n>>1)+1,0,n>>1);
	Copy((n>>1)+1,(n>>1)+1,n>>1);
	a[(n>>1)+1][(n>>1)+1]=1;
}
int main(){
	n=read(),m=read();
	dfs((1<<max(n,m))-1);
	for(int i=1;i<=(1<<n)-1;++i){
		for(int j=1;j<=(1<<m)-1;++j){
			putchar(a[i][j]+'0');
		}
		putchar('\n');
	}
	return 0;
}

