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 cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
vector<vector<int>> solve(int n, int m) {
	if (n == 0) return {vector<int>(1 << m)};
	auto g = solve(n - 1, m - 1);
	int N = 1 << n, M = 1 << m;
	vector<vector<int>> ret(N, vector<int>(M));
	for (int i = 0; i < N; ++i)
		for (int j = 0; j < M; ++j) {
			if (i < N / 2) ret[i][j] = g[i][j % (M / 2)];
			else if (j < M / 2) ret[i][j] = g[N - i - 1][j];
			else ret[i][j] = 1 - g[N - i - 1][j - M / 2];
		}
	return ret;
}
int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	vector<vector<int>> g;
	if (n <= m) g = solve(n, m);
	else {
		auto h = solve(m, n);
		g = vector<vector<int>>(1 << n, vector<int>(1 << m));
		for (int i = 0; i < h.size(); ++i)
			for (int j = 0; j < h[i].size(); ++j)
				g[j][i] = h[i][j];
	}
	for (int i = 1; i < (1 << n); ++i) {
		for (int j = 1; j < (1 << m); ++j) {
			int ans = g[i][j] ^ g[i - 1][j] ^ g[i][j - 1] ^ g[i - 1][j - 1];
			printf("%d", ans);
		}
		printf("\n");
	}
}
