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
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#endif
//#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long ll;
int n, m;
const int maxP = 1040;
int a[maxP][maxP];
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
//    freopen("input.txt", "r", stdin);
    cin >> n >> m;
    int x = (1 << n);
    int b = (1 << m);
    if (n <= m) {
        for (int i = 0; i < x; i++) {
            for (int j = 0; j < b; j++) {
                int p = (j >> (m - n));
                ::a[i][j] = __builtin_popcount((i & p)) & 1;
            }
        }
    }
    else {
        for (int i = 0; i < x; i++) {
            for (int j = 0; j < b; j++) {
                int p = (i >> (n - m));
                ::a[i][j] = (__builtin_popcount((j & p))) & 1;
            }
        }
    }



    for (int i = 1; i < (1 << n); i++) {
        if (a[i][0] == 1) {
            for (int j = 0; j < (1 << m); j++) {
                a[i][j] ^= 1;
            }
        }
    }
    for (int i = 1; i < (1 << m); i++) {
        if (a[0][i] == 1) {
            for (int j = 0; j < (1 << n); j++) {
                a[j][i] ^= 1;
            }
        }
    }
//    for (int bit1 = 0; bit1 < (1 << n); bit1++) {
//        for (int bit2 = bit1 + 1; bit2 < (1 << n); bit2++) {
//            int cnt0 = 0;
//            int cnt1 = 0;
//            for (int j = 0; j < (1 << m); j++) {
//                int p = a[bit1][j] ^ a[bit2][j];
//                if (p == 0) cnt0++;
//                else cnt1++;
//            }
//            assert(cnt0 == cnt1);
//        }
//    }
    for (int i = 1; i < (1 << n); i++) {
        for (int j = 1; j < (1 << m); j++) {
            cout << (a[i][j] ^ a[i - 1][j] ^ a[i][j - 1] ^ a[i - 1][j - 1]);
        }
        cout << '\n';
    }
    return 0;
}