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>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;

constexpr ll Mod = 998244353;
constexpr ll mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;

template <typename T> constexpr bool chmax(T &a, const T &b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T &b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    int N, M;
    cin >> N >> M;
    ll K = max(N, M);

    vector<vector<int>> x(1 << K, vector<int>(1 << K, -1));
    x[0][0] = x[0][1] = x[1][0] = 0;
    x[1][1] = 1;

    FOR(k, 1, K) {
        int z = 1 << k;
        rep(i, z) rep(j, z) {
            x[i + z][j] = x[i][j + z] = x[i][j];
            x[i + z][j + z] = 1 - x[i][j];
        }
    }

    rep(i, 1 << K) for(int j = 1 << K; --j;) x[i][j] ^= x[i][j - 1];
    rep(j, 1 << K) for(int i = 1 << K; --i;) x[i][j] ^= x[i - 1][j];

    FOR(i, 1, 1 << N) {
        FOR(j, 1, 1 << M) cout << x[i][j];
        cout << endl;
    }

    return 0;
}