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>
#define int long long
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define F first
#define S second

using namespace std;
using P = pair<int,int>;
using ivec = vector<int>;
using Graph = vector<vector<int>>;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }

int ans[1100][1100];
int a[1100][1100];

int bi(int x){
  if(x==0)return 0;
  if(x%2==0)return bi(x/2);
  return 1-bi(x/2);
}
signed main(){
  int n,m;cin>>n>>m;
  int N=max(n,m);
  rep(i,1<<N){
    rep(j,1<<N){
      a[i][j]=bi(i&j);
    }
  }
  
  for(int i=1;i<(1<<N);i++)
    for(int j=1;j<(1<<N);j++)
      ans[i][j]=(a[i][j]+a[i][j-1]+a[i-1][j]+a[i-1][j-1])%2;
  
  for(int i=1;i<(1<<n);i++){
    for(int j=1;j<(1<<m);j++){
      cout<<ans[i][j];
    }
    cout<<endl;
  }
  
  return 0;
}
