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 all(s) s.begin(), s.end()
#define pb push_back
#define ii pair<int, int>
#define x first
#define y second
#define bit(x, y) ((x >> y) & 1)
#define sz(s) ((int) (s).size())

int ans[2000][2000];

void solve(int x, int y, int u, int v) {
    if (u == 0 || v == 0) return;
    ans[x + (1 << (u - 1))][y + (1 << (v - 1))] = 1;
    solve(x, y, u - 1, v - 1);
    solve(x + (1 << (u - 1)), y, u - 1, v - 1);
    solve(x, y + (1 << (v - 1)), u - 1, v - 1);
    solve(x + (1 << (u - 1)), y + (1 << (v - 1)), u - 1, v - 1);
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); 
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    solve(0, 0, 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;
}