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>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <fstream>
#include <cassert>
#include <cstring>
#include <unordered_set>
#include <unordered_map>
#include <numeric>
#include <ctime>
#include <bitset>
#include <complex>
#include <random>

using namespace std;

const int N = (1 << 10) + 1;

int ans[N][N];

void run(int len, int i, int j) {
	if (len == 1) {
		ans[i][j] = 1;
	} else {
		for (int x = 0; x < 2; x++) {
			for (int y = 0; y < 2; y++) {
				run(len / 2, i + x * (len / 2 + 1), j + y * (len / 2 + 1));
			}
		}
		ans[i + len / 2][j + len / 2] = 1;
	}
}

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	int n, m;
	cin >> n >> m;
	n = (1 << n) - 1;
	m = (1 << m) - 1;
	int t = max(n, m);
	run(t, 0, 0);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			cout << ans[i][j];
		}
		cout << '\n';
	}
}
