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 int LL;
typedef long long int ll;
typedef pair<long long int, long long int> pii;
typedef pair<double, double> pdd;

#define SORT(c) sort((c).begin(),(c).end())
#define BACKSORT(c) sort((c).begin(),(c).end(),std::greater<LL>())
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SP << " " <<

LL mod = 1000000007;

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);

  LL NN,MM;
  cin >> NN >> MM;

  LL N = (1 << NN) -1;
  LL M = (1 << MM) - 1;

  vector<vector<LL>> vec(N, vector<LL>(M, 0));

  // LL bit = 0b101010101010101010101;

  LL num = 0;
  REP(i, N)
  {
    REP(j,M){
      vec[i][j] = 0;
      LL ii=i;
      LL jj = j;
      while (true)
      {
        if (ii % 2 == 0 && jj % 2 == 0)
        {
          vec[i][j] = 1;
          break;
        }
        if(ii%2==1&&jj%2==1){
          ii /= 2;
          jj /= 2;
          continue;
        }
        break;
      }
      num++;
    }
  }

    // LL sum = 0;
    // LL summ = 0;

    // REP(i,N){
    //   FOR(ii,i+1,N+1){
    //     REP(j,M){
    //       FOR(jj,j+1,M+1){
    //         LL cnt = 0;
    //         FOR(iii, i, ii)
    //         {
    //           FOR(jjj,j,jj){
    //             cnt += vec[iii][jjj];
    //           }
    //         }

    //         summ++;
    //         if (cnt % 2 == 1)
    //         {
    //           sum++;
    //         }
    //       }
    //     }
    //   }
    // }
    
    // cout << sum << endl;
    // cout << summ << endl;
    REP(i,N){
      REP(j,M){
        cout << vec[i][j];
      }
      cout << endl;
    }

}
