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 max_n = 1111, inf = 1000111222;

int n, m, a[max_n][max_n];
bool rev;
bitset<max_n> b[max_n];

long long solve() {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            b[i][j] = a[i - 1][j - 1] ^ b[i][j - 1];
        }
        for (int j = 0; j <= m; ++j) {
            b[i][j] = b[i][j] ^ b[i - 1][j];
        }
    }
    long long res = 0;
    for (int i = 0; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            int c1 = (b[i] ^ b[j]).count();
            int c2 = m + 1 - c1;
            res += c1 * c2;
        }
    }
    return res;
}

void proc(int lx, int rx, int ly, int ry) {
    if (lx > rx || ly > ry) {
        return;
    }
    if (lx == rx && ly == ry) {
        return;
    }
    int mx = (lx + rx) / 2;
    int my = (ly + ry) / 2;
    a[mx][my] = 0;
    proc(lx, mx - 1, ly, my - 1);
    proc(lx, mx - 1, my + 1, ry);
    proc(mx + 1, rx, ly, my - 1);
    proc(mx + 1, rx, my + 1, ry);
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    cin >> n >> m;
    if (n > m) {
        swap(n, m);
        rev = 1;
    }
    n = 1 << n;
    m = 1 << m;
    --n;
    --m;
    if (n == 1 && 0) {
        a[0][m / 2] = 1;
    } else {
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                a[i][j] = 1;
            }
        }
        proc(0, n - 1, 0, m - 1);
        /*a[n / 4][m / 4] = 0;
        a[n - 1 - n / 4][m / 4] = 0;
        a[n / 4][m - 1 - m / 4] = 0;
        a[n - 1 - n / 4][m - 1 - m / 4] = 0;
        a[n / 2][m / 2] = 0;*/
    }
    //cout << solve() << endl;
    if (!rev) {
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                cout << a[i][j];
            }
            cout << endl;
        }
    } else {
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                cout << a[j][i];
            }
            cout << endl;
        }
    }
    return 0;
}
