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 rep(i,n) for(int i=0;i<n;i++)
#define int long long

int ans[1024][1024];

void construct() {
	ans[1][1] = 1;
	for (int k = 1; k <= 9;k++) {
		rep(i, 1 << k) {
			rep(j, 1 << k) {
				ans[i][j + (1 << k)] = ans[i + (1 << k)][j] = ans[i][j];
				ans[i + (1 << k)][j + (1 << k)] = 1 ^ ans[i][j];
			}
		}
	}
}

signed main() {
	int n, m; cin >> n >> m;
	construct();
	rep(i, (1 << n)-1) {
		rep(j, (1 << m)-1) {
			cout << (ans[i+1][j+1]^ans[i][j+1]^ans[i+1][j]^ans[i][j]);
		}cout << endl;
	}
}