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;
const int N=1<<10;
int n,m;
char a[N][N];
void solve(int x)
{
    int now=1;x=(1<<x)-1;
    a[1][1]='1';
    while(now<x)
    {
        int nx=now*2+1;
        for(int i=1;i<=now;i++)
            for(int j=1;j<=now;j++)
            a[i+now+1][j+now+1]=a[i+now+1][j]=a[i][j+now+1]=a[i][j];
        for(int j=1;j<=nx;j++)
            a[now+1][j]='1';
        now=nx;
    }
}
int main()
{
    memset(a,'0',sizeof(a));
    scanf("%d%d",&n,&m);
    if(n>m)
    {
        solve(n);
        for(int i=1;i<=(1<<n)-1;i++)
        {
            for(int j=1;j<=(1<<m)-1;j++)
                putchar(a[i][j]);
            putchar('\n');
        }
    }
    else
    {
        solve(m);
        for(int i=1;i<=(1<<n)-1;i++)
        {
            for(int j=1;j<=(1<<m)-1;j++)
                putchar(a[j][i]);
            putchar('\n');
        }
    }
}
