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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
const int N = 1505; 
using namespace std;

int n, m, cnt[N], sum[N][N]; 

template < typename T >
inline T read()
{
	T x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * w; 
}

int main()
{
	n = read <int> (), m = read <int> (), n = (1 << n) - 1, m = (1 << m) - 1; 
	for(int i = 1; i <= n; i++) cnt[i] = cnt[i >> 1] + (i & 1); 
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++) sum[i][j] = cnt[i & j] & 1; 
	for(int i = 1; i <= n; i++)
	{
		for(int j = 1; j <= m; j++)
			printf("%d", sum[i][j] ^ sum[i - 1][j] ^ sum[i][j - 1] ^ sum[i - 1][j - 1]);
		puts(""); 
	}
	return 0; 
}
