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 <algorithm>
#include <bitset>
#include <cassert>
#include <iostream>
#include <vector>
using namespace std;

template<typename T> ostream& operator<<(ostream &os, const vector<T> &v) { os << "{"; for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) { if (vi != v.begin()) os << ", "; os << *vi; } os << "}"; return os; }
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { os << '(' << p.first << ", " << p.second << ')'; return os; }

void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }

#ifdef NEAL
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif


const int MAX = 1024;

bitset<MAX> grid[MAX];

// https://en.wikipedia.org/wiki/Hadamard_matrix
void generate(int size) {
    if (size == 1) {
        grid[0][0] = 0;
        return;
    }

    generate(size / 2);

    for (int i = 0; i < size / 2; i++)
        for (int j = 0; j < size / 2; j++) {
            grid[size / 2 + i][j] = grid[i][j];
            grid[i][size / 2 + j] = grid[i][j];
            grid[size / 2 + i][size / 2 + j] = !grid[i][j];
        }
}

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

    int N, M;
    cin >> N >> M;
    int NN = (1 << N) - 1, MM = (1 << M) - 1;
    generate(max(NN, MM) + 1);

    for (int i = 0; i < NN; i++) {
        for (int j = 0; j < MM; j++) {
            int ans = (grid[i][j] ^ grid[i][j + 1] ^ grid[i + 1][j] ^ grid[i + 1][j + 1]);
            cout << ans;
        }

        cout << '\n';
    }
}
