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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<int(n);i++)
#define N 200020
int a[1024][1024];
int b[1023][1023];


int main() {
	int n, m;
	int nn, mm;
	a[0][0] = 0;
	int x = 1;
	f(ii, 10) {
		f(i, x) {
			f(j, x) {
				a[x + i][j] = a[i][j];
				a[i][x + j] = a[i][j];
				a[x + i][x + j] = 1 - a[i][j];
			}
		}
					x = x * 2;
	}
	f(i, 1023) {
		f(j, 1023) {
			b[i][j] = (a[i][j] + a[i][j + 1] + a[i + 1][j] + a[i + 1][j + 1]) % 2;
		}
	}
	scanf("%d %d", &nn, &mm);
	n = 1;
	m = 1;
	f(i, nn)n = n * 2;
	f(i, mm)m = m * 2;
	n--;
	m--;
	f(i, n) {
		f(j, m) {
			printf("%d", b[i][j]);
		}
		printf("\n");
	}


	return 0;
}