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 int long long
int r(int x)
{
    int ans=1;
    while(x%ans==0)
    {
        ans*=2;
    }
    ans/=2;
    return ans;
}
int32_t main()
{
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n,m;
    cin>>n>>m;
    bool h=true;
    if(n>m)
    {
        h=false;
        swap(n,m);
    }
    int sz1=(1<<n)-1;
    int sz2=(1<<m)-1;
    int res[sz1][sz2];
    int pr[sz1+1][sz2];
    for(int i=0;i<sz1;++i)
    {
        pr[i][0]=0;
        for(int j=0;j<sz2;++j)
        {
            int x=r(i+1); // power
            if((j+1) & x)
            {
                res[i][j]=(pr[i][j]+1)%2;
                pr[i][j+1]=1;
            }
            else
            {
                res[i][j]=pr[i][j];
                pr[i][j+1]=0;
            }
        }
    }
    if(h)
    {
        for(int i=0;i<sz1;++i)
        {
            for(int j=0;j<sz2;++j)
            {
                cout<<res[i][j];
            }
            cout<<endl;
        }
    }
    else
    {
        for(int i=0;i<sz2;++i)
        {
            for(int j=0;j<sz1;++j)
            {
                cout<<res[j][i];
            }
            cout<<endl;
        }
    }
    return 0;
}
