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<cstdio>
using namespace std;
int n,m,Sum[1024][1024],a[1024][1024],Cnt[1024];
int read()
{
  char c=0;
  int sum=0;
  while (c<'0'||c>'9')
    c=getchar();
  while ('0'<=c&&c<='9')
    {
      sum=sum*10+c-'0';
      c=getchar();
    }
  return sum;
}
int lowbit(int x)
{
  return x&(-x);
}
int main()
{
  n=read(),m=read();
  for (int i=1;i<=(1<<max(n,m))-1;++i)
    Cnt[i]=Cnt[i-lowbit(i)]^1;
  for (int i=1;i<=(1<<n)-1;++i)
    for (int j=1;j<=(1<<m)-1;++j)
      Sum[i][j]=Cnt[i&j];
  for (int i=1;i<=(1<<n)-1;++i)
    for (int j=1;j<=(1<<m)-1;++j)
      a[i][j]=Sum[i][j]^Sum[i-1][j]^Sum[i][j-1]^Sum[i-1][j-1];
  for (int i=1;i<=(1<<n)-1;++i)
    {
      for (int j=1;j<=(1<<m)-1;++j)
        printf("%d",a[i][j]);
       puts("");
    }
  return 0;
}