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;
#define x first
#define y second
#define pii pair<int , int>
#define mp make_pair
#define pb push_back
#define sz(a) (int)(a.size())
const int INF = 1000 * 1000 * 1000 + 7;
const ll LINF = INF * (ll)INF;
const int mod = INF;
const int MAX = 1 << 25;
int a[1024][1024];
int b[1024][1024];
int n, m;
void print()
{
	for(int i = 1; i < (1 << n); i++)
		for(int j = 1; j < (1 << m); j++)
			b[i][j] = (a[i - 1][j] ^ a[i][j - 1] ^ a[i - 1][j - 1]) ^ a[i][j];
	for(int i = 1; i < (1 << n); i++)
	{
		for(int j = 1; j < (1 << m); j++)
			cout << b[i][j];
		cout << endl;
	}
	cout << endl;
}
void solve(int n)
{
	for(int i = 1; i <= n; i++)
	{
		int pw2 = (1 << i) , pw22 = (1 << (i - 1));
		for(int j = 0; j < pw2; j++)
			for(int k = 0; k < pw2; k++)
				if(j >= pw22 && k >= pw22)
					a[j][k] = a[j - pw22][k - pw22] ^ 1;
				else if(j >= pw22)
					a[j][k] = a[j - pw22][k];
				else if(k >= pw22)
					a[j][k] = a[j][k - pw22];
	}
}
int main()
{
	ios_base::sync_with_stdio(0);
	cin >> n >> m;
	solve(10);
	print();
}
