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;

typedef long long LL;

const int N=10;

int n,m;

void into(){
  scanf("%d%d",&n,&m);
}

int ans[(1<<N)+9][(1<<N)+9];

void Dfs_ans(int x,int y,int n){
  if (!n) return;
  int t=1<<n-1;
  ans[x+t][y+t]=1;
  Dfs_ans(x,y,n-1);
  Dfs_ans(x,y+t,n-1);
  Dfs_ans(x+t,y,n-1);
  Dfs_ans(x+t,y+t,n-1);
}

void work(){
  Dfs_ans(0,0,max(n,m));
}

void outo(){
  for (int i=1;i<1<<n;++i){
	for (int j=1;j<1<<m;++j)
      putchar(ans[i][j]+'0');
    puts("");
  }
}

int main(){
  into();
  work();
  outo();
  return 0;
}