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;
#define int long long
const int MOD = 1000000007;
int bcnt(int a) {
	int res = 0;
	while (a > 0) {
		res += (a & 1);
		a /= 2;
	}
	return res;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N, M;
	cin >> N >> M;
	bool flip = false;
	if (N > M) {
		flip = true;
		swap(N, M);
	}
	int X = 1 << N;
	int Y = 1 << M;
	vector<vector<int> > A(X, vector<int>(Y));
	// for (int i = 0; i < X; i++) {
	// 	int t = 1 << (M + 1 - (i % (M + 1)));
	// 	for (int j = 0; j < Y; j++) {
	// 		if (j % t >= t / 2) {
	// 			A[i][j] = 1;
	// 		}
	// 	}
	// }
	for (int i = 0; i < X; i++) {
		for (int j = 0; j < Y; j++) {
			A[i][j] = bcnt(i & j) % 2;
		}
	}

	// for (int i = 0; i < X; i++) {
	// 	for (int j = 0; j < Y; j++) {
	// 		cerr << A[i][j] << " ";
	// 	}
	// 	cerr << endl;
	// }
	vector<vector<int> > res(X - 1, vector<int>(Y - 1));
	for (int i = 0; i < X - 1; i++) {
		for (int j = 0; j < Y - 1; j++) {
			res[i][j] = (A[i][j] ^ A[i + 1][j] ^ A[i][j + 1] ^ A[i + 1][j + 1]);
		}
	}
	if (flip) {
		for (int j = 0; j < Y - 1; j++) {
			bool f = true;
			for (int i = 0; i < X - 1; i++) {
				cout << res[i][j];
			}
			cout << endl;
		}
	} else {
		for (int i = 0; i < X - 1; i++) {
			bool f = true;
			for (int j = 0; j < Y - 1; j++) {
				cout << res[i][j];
			}
			cout << endl;
		}
	}
}