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;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

int n, m;
int swapped = 0;
int arr[1050][1050];

int main() {
	scanf("%d%d",&n,&m);
	if (n<m) swap(n,m), swapped=1;
	for (int i=0;i<m;i++) {
		for (int j=0;j<(1<<m);j++) {
			if (j>>i&1) {
				for (int k=0;k<(1<<n);k++) {
					if (k>>i&1) arr[k][j]^=1;
				}
			}
		}
	}
	if (swapped) swap(n,m);
	for (int i=1;i<(1<<n);i++) {
		for (int j=1;j<(1<<m);j++) {
			printf("%d",swapped?(arr[j][i]^arr[j-1][i]^arr[j][i-1]^arr[j-1][i-1]):(arr[i][j]^arr[i-1][j]^arr[i][j-1]^arr[i-1][j-1]));
		}
		puts("");
	}

	return 0;
}
