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;

int n, m, g[1024][1024] = {0};
int e[11] = {0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023};

void construct (int x) {
    if (x == 1) {
        g[x][x] = 1; return;
    }
    int y = x - 1;
    construct (y);
    g[e[y] + 1][e[y] + 1] = 1;
    for (int i = e[y] + 2; i <= e[x]; i ++)
        for (int j = 1; j <= e[y]; j ++)
            g[i][j] = g[i - e[y] - 1][j];
    
    for (int i = 1; i <= e[x]; i ++)
        for (int j = e[y] + 2; j <= e[x]; j ++)
            g[i][j] = g[i][j - e[y] - 1];
}

int main () {
    cin >> n >> m;
    construct (max (n, m));
    for (int i = 1; i <= e[n]; i ++) {
        for (int j = 1; j <= e[m]; j ++)
            cout << g[i][j];
        cout << endl;
    }
    return 0;
}