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 PB push_back
#define MP std::make_pair
#define FI first
#define SE second
#define M 1000005
#define INF 1000000000
int n,m,S[1024][1024];
int read(){
	char c=getchar();int ans=0;
	while (c<'0'||c>'9') c=getchar();
	while (c>='0'&&c<='9') ans=ans*10+c-'0',c=getchar();
	return ans;
}
void Write(int x){
	if (x<10) putchar(x+'0');
	else Write(x/10),putchar(x%10+'0');
	return;
}
int max(int x,int y){return x>y?x:y;}
void calc(int X1,int Y1,int X2,int Y2){
	if (X1==X2) return S[X1][Y1]=1,void();
	int midX=(X1+X2)/2,midY=(Y1+Y2)/2;
	calc(X1,Y1,midX-1,midY-1);
	calc(X1,midY+1,midX-1,Y2);
	calc(midX+1,Y1,X2,midY-1);
	calc(midX+1,midY+1,X2,Y2);
	return S[midX][midY]=1,void();
}
int main(){
	n=(1<<read())-1,m=(1<<read())-1,calc(1,1,max(n,m),max(n,m));
	for (int i=1;i<=n;i++,putchar('\n'))
		for (int j=1;j<=m;j++) putchar(S[i][j]+'0');
	return 0;
}
