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>
using namespace std;

#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef complex<double> point;
typedef double DOUBLE;
#define xx real()
#define yy imag()

#define REP(i, a, b) for(int i = (a); i < (int)(b); i++)
#define REPN(i, a, b) for(int i = (a); i <= (int)(b); i++)
#define FA(it, x) for(__typeof((x).begin()) it = (x).begin(); it != (x).end(); it++)
#define SZ(x) (int)(x).size()
#define BE(x) (x).begin(), (x).end()
#define SORT(x) sort(BE(x))
#define _1 first
#define _2 second

#define x1 gray_cat_x1
#define y1 gray_cat_y1

template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define file "I1"

const double EPS = 1e-9;
const double PI = acos(-1.);
const ll LL_INF = 1e17 + 10;
const int INF = 1e9 + 10;
const ll MOD = 1e9 + 7;

const int MAXN = 1 << 10;

int g[MAXN][MAXN];

void rec_build(int t, int l, int r) {
	if (t == 1) {
		g[l][r] = 1;
		return;
	}
	int t1 = t / 2;
	rec_build(t1, l, r);
	rec_build(t1, l + t1 + 1, r);
	rec_build(t1, l, r + t1 + 1);
	rec_build(t1, l + t1 + 1, r + t1 + 1);
	g[l + t1][r + t1] = 1;
}

void solve(){
	int n, m;
	scanf("%d%d", &n, &m);
	n = (1 << n) - 1, m = (1 << m) - 1;
	int t = min(n, m);
	rec_build(t, 0, 0);
	
	if (t < n) {
		for(int i = t + 1; i < n; i += t + 1) {
			rec_build(t, i, 0);
		}
	}
	
	if (t < m) {
		for(int i = t + 1; i < m; i += t + 1) {
			rec_build(t, 0, i);
		}
	}
	
	int LIMI = n, LIMJ = m;
	/*int mx = -1;

	//g[3][3] = 0;
	int cur = 0;
	REP(i1, 0, LIMI) {
		REP(i2, i1, LIMI) {
			REP(j1, 0, LIMJ) {
				REP(j2, j1, LIMJ) {
					int cnt = 0;
					REPN(i, i1, i2) {
						REPN(j, j1, j2) {
							cnt += g[i][j];
						}
					}
					if (cnt & 1) {
						cur++;
					}
				}
			}
		}
	}

	printf("%d\n", cur);*/
	
	REP(i, 0, LIMI) {
		REP(j, 0, LIMJ) {
			printf("%d", g[i][j]);
		}
		putchar('\n');
	}
}

int main(){

    //freopen(file".in", "r", stdin); freopen(file".out", "w", stdout);
    int t = 1;
    //cin >> t;
    while(t--){
        solve();    
    }
}
