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>
#include <algorithm>
#include <cstring>
#include <map>
#include <vector>
using namespace std;
typedef long long LL;

LL f[2055][2055] = {0};
LL n,m;

void solve(LL x,LL y,LL d){
	if(d == 0){
		f[x][y] = 1;
		return;
	}
	solve(x,y,d - 1);
	solve(x + (1LL << d),y,d - 1);
	solve(x,y + (1LL << d),d - 1);
	solve(x + (1LL << d),y + (1LL << d),d - 1);
	f[x + (1LL << d) - 1][y + (1LL << d) - 1] = 1;
}

int main(){
	cin >> n >> m;
	solve(1,1,10);
	for(LL i = 1;i < (1LL << n);i ++){
		for(LL j = 1;j < (1LL << m);j ++){
			cout << f[i][j];
		}
		cout << endl;
	}
	return 0;
}