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>
 
#define f first
#define s second
#define pb push_back
#define mp make_pair
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
 
const int N = (int)2e5 + 123, inf = 1e9, mod = 1e9 + 7;
const ll INF = 1e18;
 
int n, m, a[1100][1100];
void solve(int n, int m){
	if(n > m){
		solve(n / 2, m);
		for(int i = 0; i < n / 2; i++)
			for(int j = 0; j < m; j++)
				a[i + n / 2 + 1][j] = a[i][j];
		for(int j = 0; j < m; j++)
			a[n / 2][j] = 0;
		return;
	}
	if(m > n){
		solve(n, m / 2);
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m / 2; j++)
				a[i][j + m / 2 + 1] = a[i][j];
		for(int i = 0; i < n; i++)
			a[i][m / 2] = 0;	
		return;
	}
	if(n == 1){
		a[0][0] = 1;
		return;
	}
	solve(n / 2, m / 2);
	for(int i = 0; i < n / 2; i++)
		for(int j = 0; j < n / 2; j++)
			a[i + n / 2 + 1][j + n / 2 + 1] = a[i][j + n / 2 + 1] = a[i + n / 2 + 1][j] = a[i][j];
	for(int i = 0; i < n; i++)
		a[i][n / 2] = a[n / 2][i] = 0;
	a[n / 2][m / 2] = 1;	
}
int main()
{
	cin >> n >> m;
	solve((1 << n) - 1, (1 << m) - 1);
	for(int i = 0; i < ((1 << n) - 1); i++){
		for(int j = 0; j < ((1 << m) - 1); j++)
			printf("%d", a[i][j]);
		printf("\n");
	}
}
