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<cstdio>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
int a[1100][1100];
inline void solve(int n)
{
	if(n==1){a[2][2]=1;return ;}
	else
	{
		solve(n-1);
		int ul=1<<(n-1),uf=1<<n;
		for(int i=1;i<=ul;i++)
		{
			for(int j=ul+1;j<=uf;j++)a[i][j]=a[i][j-ul];
		}
		for(int i=ul+1;i<=uf;i++)
		{
			for(int j=1;j<=ul;j++)a[i][j]=a[i-ul][j];
			for(int j=ul+1;j<=uf;j++)a[i][j]=a[i][j-ul]^1;
		}
	}
}
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    int n=read(),m=read(),T=max(n,m);
    solve(T);
    n=1<<n,m=1<<m,T=1<<T;
    for(int i=T;i>=2;i--)
    {
    	for(int j=T;j>=2;j--)a[i][j]^=a[i-1][j]^a[i][j-1]^a[i-1][j-1];
    }
    for(int i=2;i<=n;i++)
    {
    	for(int j=2;j<=m;j++)write(a[i][j]);
    	puts("");
    }
    return 0;
}