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 endl enjoy_codeforces
int main(){
    std::cin.tie(nullptr);std::ios_base::sync_with_stdio(false);
    int n,m;std::cin>>n>>m;
    int N=(1<<n)-1,M=(1<<m)-1;
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            int I=0;for(int k=i+1;!(k&1);k>>=1)I++;
            int J=0;for(int k=j+1;!(k&1);k>>=1)J++;
            std::cout<<(I+J==std::max(n,m)?0:1);
        }
        std::cout<<'\n';
    }
}
