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>
// iostream is too mainstream
#include <cstdio>
// bitch please
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <list>
#include <cmath>
#include <iomanip>
#include <time.h>
#define dibs reserve
#define OVER9000 1234567890
#define ALL_THE(CAKE,LIE) for(auto LIE =CAKE.begin(); LIE != CAKE.end(); LIE++)
#define tisic 47
#define soclose 1e-8
#define chocolate win
// so much chocolate
#define patkan 9
#define ff first
#define ss second
#define abs(x) (((x) < 0)?-(x):(x))
#define uint unsigned int
#define dbl long double
#define pi 3.14159265358979323846
using namespace std;
// mylittledoge

using cat = long long;

#ifdef DONLINE_JUDGE
	// palindromic tree is better than splay tree!
	#define lld I64d
#endif

int main() {
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout << fixed << setprecision(10);
	int N, M;
	cin >> N >> M;
	bool sw = false;
	if(N < M) swap(N, M), sw = true;
	vector< vector<char> > V(1<<N, vector<char>(1<<M, 0));
	for(int i = 0; i < (1<<M); i++)
		for(int j = 0; j < (1<<M); j++)
			V[i][j] = __builtin_popcount(i&j) & 1;
	for(int i = (1<<M); i < (1<<N); i++)
		for(int j = 0; j < (1<<M); j++)
			V[i][j] = V[i-(1<<M)][j];
	if(sw) {
		vector< vector<char> > VT(1<<M, vector<char>(1<<N, 0));
		for(int i = 0; i < (1<<N); i++)
			for(int j = 0; j < (1<<M); j++)
				VT[j][i] = V[i][j];
		V = VT;
		swap(N, M);
	}
	for(int i = (1<<N)-1; i > 0; i--)
		for(int j = (1<<M)-1; j > 0; j--)
			V[i][j] ^= V[i-1][j] ^ V[i-1][j-1] ^ V[i][j-1];
	for(int i = 1; i < (1<<N); i++) {
		for(int j = 1; j < (1<<M); j++) cout << (int)V[i][j];
		cout << "\n";
	}
	return 0;
}

// look at my code
// my code is amazing
