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>
using namespace std;

int N, M;
int P[1 << 11][1 << 11];

void solve(int lx, int rx, int ly, int ry) {
	P[(lx + rx) / 2][(ly + ry) / 2] = 1;
	if (lx == rx && ly == ry) return;
	
	int mx = (lx + rx) / 2, my = (ly + ry) / 2;
	solve(lx, mx - 1, ly, my - 1);
	solve(lx, mx - 1, my + 1, ry);
	solve(mx + 1, rx, ly, my - 1);
	solve(mx + 1, rx, my + 1, ry);
}

int main() {
	cin >> N >> M;
	solve(1, 1023, 1, 1023);

	for (int i = 1; i <= (1 << N) - 1; i++) {
		for (int j = 1; j <= (1 << M) - 1; j++) cout << P[i][j];
		cout << endl;
	}
	return 0;
}