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"
using namespace std;

int main() {
	int N, M;
	cin >> N >> M;
	int X = max(N, M);
	int Z = 1 << X;
	vector<vector<int> > V(Z, vector<int>(Z));
	V[0][0] = 0;
	for (int i = 0; i < X; i++) {
		for (int j = 0; j < 1 << i; j++) {
			for (int k = 0; k < 1 << i; k++) {
				V[j + (1 << i)][k] = V[j][k + (1 << i)] = V[j][k];
				V[j + (1 << i)][k + (1 << i)] = 1 - V[j][k];
			}
		}
	}
	vector<vector<int> > ANS(Z - 1, vector<int>(Z - 1, 0));
	for (int i = 0; i < Z - 1; i++) {
		for (int j = 0; j < Z - 1; j++) {
			ANS[i][j] = (V[i][j] + V[i][j + 1] + V[i + 1][j] + V[i + 1][j + 1]) % 2;
		}
	}
	for (int i = 0; i < (1 << N) - 1; i++) {
		for (int j = 0; j < (1 << M) - 1; j++) {
			cout << ANS[i][j];
		}
		cout << endl;
	}
}