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;

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep3(i,m,n) for(int (i)=m;(i)<=(n);(i)++)
#define rep3rev(i,m,n) for(int (i)=m;(i)>=(n);(i)--)
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define eb emplace_back

using ll = long long;
using vll = vector<ll>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using P = pair<int, int>;

void Main(){
    int n, m; cin >> n >> m;
    bool rev = false;
    if(n <= m) {
        swap(n, m);
        rev = true;
    }
    
    int N = 1 << n;
    int M = 1 << m;
    vvi sum(N, vi(N, 0)), sum_i(N, vi(N, 0)), ans(N, vi(N, 0));
    
    rep(i, N){
        rep(j, N){
            sum[i][j] = __builtin_popcount(i & j) % 2; 
        }
    }

    rep3(i, 1, N-1){
        rep(j, N){
            sum_i[i][j] = (sum[i][j] + sum[i-1][j]) % 2;
        }
    }

    rep(i, N){
        rep3(j, 1, N-1){
            ans[i][j] = (sum_i[i][j] + sum_i[i][j-1]) % 2;
        }
    }
    
    /*
    rep(i, N){
        rep(j, M){
            cout << sum[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
    
    rep(i, N){
        rep(j, M){
            cout << sum_i[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
    */
    
    if(rev) swap(N, M);
    
    rep3(i, 1, N-1){
        rep3(j, 1, M-1){
            cout << ans[i][j];
        }
        cout << endl;
    }
    return;
}

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
    return 0;
}