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 <vector>

using ll = long long int;

using namespace std;


ll N, M;
ll N2, M2;


int main() {
  cin >> N >> M;
  N2 = (1 << N) - 1;
  M2 = (1 << M) - 1;

  ll tz[114514] = { 0 };

  for (ll i = 1; i < 114514; i++) {
    ll count = 0;
    ll x = i;
    while ((x & 1) == 0) {
      x = x >> 1;
      count++;
    }

    tz[i] = count;
  }


  for (ll i = 1; i <= N2; i++) {
    for (ll j = 1; j <= M2; j++) {
      if (tz[i] == tz[j]) {
        cout << '1';
      } else {
        cout << '0';
      }
    }
    cout << endl;
  }

  return 0;
}
