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 <cstdio>

using namespace std;

int a[1 << 10][1 << 10];
int ans[1 << 10][1 << 10];

int main() {
    int n, m, i, j, k;
    
    scanf("%d %d", &n, &m);
    
    a[1][1] = 1;
    
    for (i = 1; i < 10; i++) {
        int c = (1 << i);
        for (j = 0; j < c; j++) {
            for (k = 0; k < c; k++) {
                a[j + c][k] = a[j][k];
                a[j][k + c] = a[j][k];
                a[j + c][k + c] = 1 - a[j][k];
            }
        }
    }
    
    for (i = 0; i < (1 << n); i++) {
        int p = 0;
        for (j = 0; j < (1 << m); j++) {
            if (i == 0 || j == 0) {
                ans[i][j] = a[i][j];
            } else {
                ans[i][j] = (a[i - 1][j] ^ p ^ a[i][j]);
                p ^= ans[i][j];
            }
        }
    }
    
    for (i = 1; i < (1 << n); i++) {
        for (j = 1; j < (1 << m); j++) {
            printf("%d", ans[i][j]);
        }
        puts("");
    }
    
    return 0;
}
