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 F first
#define S second
#define pb push_back
using namespace std;
typedef long long ll;
int f[2222][2222],g[2222][2222],d[2222],n,m;
int main(){
    cin >> n >> m;
    for (int i=0;i<2222;i++) d[i]=d[i/2]+(i%2);
    n=1<<n; m=1<<m;
    for (int i=0;i<n;i++)
        for (int j=0;j<m;j++)
            f[i][j]=d[i&j]&1;
    for (int i=1;i<n;i++)
        for (int j=1;j<m;j++)
            g[i][j]=f[i][j]^f[i][j-1]^f[i-1][j]^f[i-1][j-1];
    for (int i=1;i<n;i++){
        for (int j=1;j<m;j++) cout<<g[i][j];
        puts("");
    }
}