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;

const int N = (int)2e3;

int dat[N+5][N+5];

int fill(int x1, int x2, int y1, int y2) {
    int x_mid = (x1 + x2) / 2;
    int y_mid = (y1 + y2) / 2;

    dat[x_mid][y_mid] = 1;

    if (x2-x1 < y2-y1) {
        fill(x1,x2, y1, y_mid-1);
        fill(x1,x2, y_mid,y_mid);
        fill(x1,x2, y_mid+1, y2);
    } else if (x2-x1 > y2-y1) {
        fill(x1, x_mid-1, y1,y2);
        fill(x_mid,x_mid, y1,y2);
        fill(x_mid+1, x2, y1,y2);
    } else if (x1 != x2) {
        fill(x1,x_mid-1, y1,y_mid-1);
        fill(x1,x_mid-1, y_mid+1,y2);
        fill(x_mid+1,x2, y1,y_mid-1);
        fill(x_mid+1,x2, y_mid+1,y2);
    }
    return 0;
}

int main() {
    int n,m; scanf("%d %d", &n, &m);

    n = (1 << n) - 1;
    m = (1 << m) - 1;

    fill(1,n, 1,m);

    for (int i=1; i<=n; i++) {
        for (int j=1; j<=m; j++) {
            printf("%d", dat[i][j]);
        }
        printf("\n");
    }
    return 0;
}
