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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <time.h>
#define int long long
#define endl '\n'
#define INF 1000000000000000000
#define EPS 1e-10
#define PI 3.141592653589793238
#define all(a) a.begin(),a.end()
#define rep(i, n) for (int i = 0; i < (n); i++)
#define fi first
#define se second
#define pb push_back
#define double long double
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<ll, P> PP;
typedef complex<double> com;
template<class T, class S> bool chmax(T &a, const S &b) { if (a < b) { a = b; return 1; } return 0; }
template<class T, class S> bool chmin(T &a, const S &b) { if (a > b) { a = b; return 1; } return 0; }
ll gcd(ll n, ll m) { return (m ? gcd(m, n%m) : n); }
ll lcm(ll n, ll m) { return n / gcd(n, m)*m; }

vector<vector<int>> solve(int n) {
	vector<vector<int>> res(n, vector<int>(n));
	if (n == 2) {
		res[1][1] = 1;
		return res;
	}
	int m = n / 2;
	vector<vector<int>> pre = solve(m);
	rep(i, n) {
		rep(j, n) {
			if (i >= m && j >= m) {
				res[i][j] = pre[i%m][j%m] ^ 1;
			}
			else {
				res[i][j] = pre[i%m][j%m];
			}
		}
	}
	return res;
}

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(20);
	//modcalc();

	int n, m;
	cin >> n >> m;
	int x = (1 << n);
	int y = (1 << m);
	int z = min(x, y);
	vector<vector<int>> sum(x, vector<int>(y));
	vector<vector<int>> tmp = solve(z);
	rep(i, x) {
		rep(j, y) {
			sum[i][j] = tmp[i%z][j%z];
		}
	}
	vector<vector<int>> ans(x - 1, vector<int>(y - 1));
	rep(i, x - 1) {
		rep(j, y - 1) {
			ans[i][j] = (sum[i][j] + sum[i][j + 1] + sum[i + 1][j] + sum[i + 1][j + 1]) % 2;
			cout << ans[i][j];
		}
		cout << endl;
	}
}