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 <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef long long ll;

int c[1<<10][1<<10];
int cumsum[1<<10][1<<10];
int N, M;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    int n, m;
    cin >> n >> m;
    N = (1<<n)-1;
    M = (1<<m)-1;
    int k = min(n, m);
    int l = max(n, m);
    vector<string> v[10];
    v[0].push_back("00");
    v[0].push_back("01");
    for(int i = 1; i < l; i++){
        for(int j = 0; j < v[i-1].size(); j++){
            string tmp0, tmp1;
            for(int s = 0; s < v[i-1][j].size(); s++){
                if(v[i-1][j][s] == '0'){
                    tmp0 += "00";
                    tmp1 += "01";
                }else{
                    tmp0 += "11";
                    tmp1 += "10";
                }
            }
            v[i].push_back(tmp0);
            v[i].push_back(tmp1);
        }
    }
    if(n > m){
        // n = l
        for(int i = 0; i <= N; i++){
            for(int j = 0; j <= M; j++){
                cumsum[i][j] = (v[l-1][i][j]-'0');
            }
        }
    }else{
        // m == l
        for(int i = 0; i <= N; i++){
            for(int j = 0; j <= M; j++){
                cumsum[i][j] = (v[l-1][j][i]-'0');
            }
        }
    }
    // for(int i = 0; i <= N; i++){
    //     for(int j = 0; j <= M; j++){
    //         cout << cumsum[i][j];
    //     }
    //     cout << endl;
    // }
    for(int i = 0; i < N; i++){
        for(int j = 0; j < M; j++){
            c[i][j] = (cumsum[i+1][j+1]+cumsum[i][j+1]+cumsum[i+1][j]+cumsum[i][j])%2;
            cout << c[i][j];
        }
        cout << endl;
    }
}