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<stdio.h>

int main(void){
  int N,M,n,m;
  scanf("%d %d",&N,&M);
  if(N>M){
    n=M;
    m=N;
  }else{
    n=N;
    m=M;
  }
  
  int f[11];
  f[0]=1;
  int i,j;
  for(i=1;i<=10;i++){
    f[i]=f[i-1]*2;
  }
  
  if(m==1){
    if(N>M){
      for(i=1;i<=f[N]-1;i++){
        printf("1\n");
      }
    }else{
      for(i=1;i<=f[M]-1;i++){
        printf("1");
      }
    }
    return 0;
  }
  
  int x[1024][1024];
  for(j=0;j<=f[m]-1;j++){
    x[0][j]=0;
    x[1][j]=j%2;
  }
  int k=1;
  for(i=2;i<=f[n]-1;i++){
    if(i==k*2){
      for(j=0;j<=f[m]-1;j++){
        x[i][j]=(j/i)%2;
      }
      k=k*2;
    }else{
      for(j=0;j<=f[m]-1;j++){
        x[i][j]=(x[0][j]+x[i-k][j]+x[k][j])%2;
      }
    }
  }
  
  if(M>N){
    for(i=1;i<=f[n]-1;i++){
      for(j=1;j<=f[m]-1;j++){
        printf("%d",(x[i-1][j-1]+x[i-1][j]+x[i][j-1]+x[i][j])%2);
      }
      printf("\n");
    }
  }else{
    for(j=1;j<=f[m]-1;j++){
      for(i=1;i<=f[n]-1;i++){
        printf("%d",(x[i-1][j-1]+x[i-1][j]+x[i][j-1]+x[i][j])%2);
      }
      printf("\n");
    }
  }
  
  return 0;
}