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 <queue>
#include <vector>
#include <algorithm>
#include <set>
#include <cmath>
#include <tuple>
#include <cstring>
#include <map>
#include <iomanip>
#include <ctime>
#include <complex>
#include <cassert>
#include <climits>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define _ << " " <<
#define all(X) (X).begin(), (X).end()
#define len(X) (X).size()
#define Pii pair<int, int>
#define Pll pair<ll, ll>
#define Tiii tuple<int, int, int>
#define Tlll tuple<ll, ll, ll>

int a[2000][2000] = {};

void f(int x, int y, int p, int q) {
    cerr << x _ y _ p _ q << endl;
    int mx = (p + x) / 2, my = (q + y) / 2;
    a[mx][my] = 1;
    if (abs(p - x) <= 1 || abs(q - y) <= 1) return;
    f(x, y, mx - 1, my - 1);
    f(x, my + 1, mx - 1, q);
    f(mx + 1, y, p, my - 1);
    f(mx + 1, my + 1, p, q);
}


int main() {
    int n, m;
    cin >> n >> m;

    f(0, 0, (1 << n) - 2, (1 << m) - 2);
    for (int i = 0; i < (1 << n) - 1; i++) {
        for (int j = 0; j < (1 << m) - 1; j++) {
            cout << a[i][j];
        }
        cout << endl;
    }

}
