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>
#define llint long long

using namespace std;

llint h, w;
llint H, W;
llint s[1<<10][1<<10];

void calc(llint n)
{
	if(n == 0) return;
	
	calc(n-1);
	llint N = 1<<n;
	for(int y = 0; y < N; y++){
		for(int x = 0; x < N; x++){
			s[x][y] = s[x%(N/2)][y%(N/2)];
			if(x >= N/2 && y >= N/2) s[x][y] = 1 - s[x][y];
		}
	}
}

int main(void)
{
	cin >> h >> w;
	H = 1<<h, W = 1<<w;
	
	calc(10);
	
	for(int i = 1; i < H; i++){
		for(int j = 1; j < W; j++){
			cout << (s[i][j] ^ s[i-1][j] ^ s[i][j-1] ^ s[i-1][j-1]);
		}
		cout << endl;
	}
	
	return 0;
}