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;
int a[2000][2000];
void go(int x,int y,int z,int w){
    if((z-x+1)<=1||w-y+1<=1)return;
    a[(x+z)/2][(y+w)/2]=0;
    go(x,y,(x+z)/2-1,(y+w)/2-1);
    go(x,(y+w)/2+1,(x+z)/2-1,w);
    go((x+z)/2+1,y,z,(y+w)/2-1);
    go((x+z)/2+1,(y+w)/2+1,z,w);
}
int main(){
	int n,m;
  	scanf("%d %d",&n,&m);
  n=(1<<n)-1;
  m=(1<<m)-1;

    for(int i = 1;i<=n;i++){
    	for(int j=1;j<=m;j++){
        	a[i][j]=1;
        }
    }
    go(1,1,n,m);
  for(int i = 1;i<=n;i++){
  for(int j=1;j<=m;j++)printf("%d",a[i][j]);
    printf("\n");
  }
  /*for(int i = 1;i<=n;i++){
      for(int j=1;j<=m;j++){
          a[i][j]+=a[i-1][j]+a[i][j-1]-a[i-1][j-1];
      }
  }
  int sum=0;
  for(int i=0;i<n;i++){
      for(int j=0;j<m;j++){
          for(int k=i+1;k<=n;k++){
              for(int l=j+1;l<=m;l++){
                  if((a[k][l]-a[i][l]-a[k][j]+a[i][j])&1)
                    sum++;
              }
          }
      }
  }
  printf("%d\n",sum);*/
}