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
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("avx,avx2,sse,sse2,ssse3,sse4.1,sse4.2,tune=native,popcnt")
#include<bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define pb push_back
using namespace std;
const int maxn = 303030;
using vi = vector<int>;
using ll = long long;
int mat[1030][1030];
vector<vi> a;
int n, m;
ll check() {
	for(int i = 0; i < n; i++) {
		for(int j = 0; j < m; j++) {
			mat[i+1][j+1] = a[i][j] ^ mat[i][j+1];
		}
	}
	ll ans = 0;
	for(int i = 0; i <= n; i++)
		for(int j = 0; j < i; j++) {
			int o = 0, e = 0, pref = 0;
			for(int a = 0; a <= m; a++) 
					((pref ^= mat[i][a]^mat[j][a])?o:e)++;
			ans += o*1ll*e;
		}
	return ans;
}
void print() {
	for(auto i : a) {
		for(auto j : i) cout << j;
		cout << '\n';
	}
	exit(0);
}
void solve(int lx, int rx, int ly, int ry) {
	if(lx == rx && ly == ry) {
		a[lx][ly] = 1;
		return;
	}
	if(rx-lx > ry-ly) {
		int mx = (lx+rx)/2;
		for(int i = ly; i <= ry; i++)
			a[mx][i] ^= 1;
		solve(lx, mx-1, ly, ry);
		solve(mx+1, rx, ly, ry);
		return;
	}
	if(rx-lx < ry-ly) {
		int my = (ly+ry)/2;
		for(int i = lx; i <= rx; i++)
			a[i][my] ^= 1;
		solve(lx, rx, ly, my-1);
		solve(lx, rx, my+1, ry);
		return;
	}
	int mx = (lx+rx)/2;
	for(int i = ly; i <= ry; i++)
		a[mx][i] ^= 1;
	int my = (ly+ry)/2;
	for(int i = lx; i <= rx; i++)
		a[i][my] ^= 1;
	solve(lx, mx-1, ly, my-1);
	solve(lx, mx-1, my+1, ry);
	solve(mx+1, rx, ly, my-1);
	solve(mx+1, rx, my+1, ry);
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	//assert(n==m);
	n = (1<<n)-1, m = (1<<m)-1;
	a= vector<vi>(n, vi(m, 0));
	solve(0, n-1, 0, m-1);
	//cout << check() << '\n';
	print();
}
//1835008:
//2064384
//1572864:
