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<string>
#include<vector>
#include<algorithm>//sort,reverse
#include<utility>//pair
#include<queue>//queue,priority_queue
#include<deque>//double_ended queue
using namespace std;
typedef pair<int,int> P;//Pでpair<-,->を表す。
typedef long long LL; 
LL MOD=1000000007;

int main()
{ 
    int n,m;
    cin>>n>>m;
    int x=1;
    int y=1;
    for(int i=0;i<n;i++){
        x=x*2;
    }
    for(int i=0;i<m;i++){
        y=y*2;
    }
    int a[x][y];
    for(int I=1;I<x;I++){
        for(int J=1;J<y;J++){
            int i=I;
            int j=J;
            while(i%2==0&&j%2==0){i=i/2;j=j/2;}
            if((i+j)%2==0){a[I][J]=1;}
            else{a[I][J]=0;}
        }
    }
    for(int i=1;i<x;i++){
        for(int j=1;j<y;j++){
            cout<<a[i][j];
        }
        cout<<endl;
    }
    return 0;
}