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 pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " : " << x << '\n'

using namespace std;

typedef long long ll;
typedef long double ld;
typedef string str;
typedef pair<ll, ll> pll;

const ll Mod = 1000000007LL;
const int N = 2e5 + 10;
const ll Inf = 1e9 + 3;
const ll Log = 35;

vector<pll> V, V0;

ll dp[Log];

bool CMP(pll a, pll b){
	return a.S * b.F < a.F * b.S;
}
ll ans[1030][1030];

int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll n, m;
	cin >> n >> m;
	if(n <= m){
	for(int i = 1; i < (1 << n); i++){
		ll h = 0;
		for(int j = 0; j < n; j++){
			if(i >> j & 1){
				h = j;
				break;
			}
		}
		for(int j = 1; j < (1 << m); j++){
			ans[i][j] = (j % (1 << h) == 0 ? 1 : 0);
		}
	}
	} else {
		swap(n, m);
		for(int i = 1; i < (1 << n); i++){
		ll h = 0;
		for(int j = 0; j < n; j++){
			if(i >> j & 1){
				h = j;
				break;
			}
		}
		for(int j = 1; j < (1 << m); j++){
			ans[j][i] = (j % (1 << h) == 0 ? 1 : 0);
		}
		}
		swap(n, m);
	}
	for(int i = 1; i < (1 << n); i++){
		for(int j = 1; j < (1 << m); j++) cout << ans[i][j];
			cout << '\n';
	}
	return 0;
}
