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 <iostream>
#include <vector>

template <class T>
std::vector<T> vec(int len, T elem) { return std::vector<T>(len, elem); }

void solve() {
    int n, m;
    std::cin >> n >> m;

    auto ans = vec(1 << n, vec(1 << m, 0));
    for (int i = 0; i < (1 << n); ++i) {
        for (int j = 0; j < (1 << m); ++j) {
            ans[i][j] = __builtin_popcount(i & j) % 2;
        }
    }

    for (int i = 1; i < (1 << n); ++i) {
        for (int j = 1; j < (1 << m); ++j) {
            int b = (ans[i][j] + ans[i - 1][j] +
                     ans[i][j - 1] + ans[i - 1][j - 1]) %
                    2;
            std::cout << b;
        }
        std::cout << std::endl;
    }
}

int main() {
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    std::ios::sync_with_stdio(false);

    solve();

    return 0;
}
