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=2020;
int n,m;bool fl;char c[N][N],t[N][N];
void solve(int n){
    if(n==1){c[1][1]=1;return;}
    solve(n-1);int m=1<<n-1;
    for(int i=1;i<m;i++)for(int j=1;j<m;j++)
        t[i][j]=t[i][j+m]=t[i+m][j]=t[i+m][j+m]=c[i][j];
    t[m][m]=1;
    for(int i=1;i<2*m;i++)for(int j=1;j<2*m;j++)c[i][j]=t[i][j];
}
int main(){
    scanf("%d%d",&n,&m);
    if(n<m)swap(n,m),fl=1;
    int pn=1<<n,pm=1<<m;
    solve(n);for(int i=1;i<pn;i++)for(int j=1;j<pn;j++)c[i][j]+='0';
    if(fl){
        for(int i=1;i<pn;i++)for(int j=i;j<pn;j++)swap(c[i][j],c[j][i]);
        for(int i=1;i<pm;i++)printf("%s\n",c[i]+1);
    }
    else for(int i=1;i<pn;i++)c[i][pm]=0,printf("%s\n",c[i]+1);
    return 0;
}