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 <cstdio>
#include <iostream>
using namespace std;
const int N = 1025, M = 1025;
int n, m, ans[N][M];
void Solve(int L, int R, int U, int D) {
	if (L > R || U > D)
		return ;
	if (L == R && U == D) {
		ans[L][U] = 1;
		return ;
	}
	int cMid = (L + R) >> 1, rMid = (U + D) >> 1;
	ans[cMid][rMid] = 1;
	Solve(L, cMid - 1, U, rMid - 1);
	Solve(cMid + 1, R, U, rMid - 1);
	Solve(L, cMid - 1, rMid + 1, D);
	Solve(cMid + 1, R, rMid + 1, D);
}
int main() {
	scanf("%d%d", &n, &m);
	int mx = max(n, m);
	Solve(1, (1 << mx) - 1, 1, (1 << mx) - 1);
	for (int i = 1; i < (1 << n); ++i) {
		for (int j = 1; j < (1 << m); ++j)
			printf("%d", ans[i][j]);
		printf("\n");
	}
	return 0;
}