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;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/e.in", "r", stdin);
//  freopen("../IO/e.out", "w", stdout);
#else
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
#endif
  solve(std::cin, std::cout);
  return 0;
}

int mp[1 << 11][1 << 11];
void solve(std::istream &in, std::ostream &out) {
  int n, m;
  in >> n >> m;
  bool swapped = false;
  if (n < m) swap(n, m), swapped = true;
  mp[1][1] = 1;
  for (int len = 1; len < n; ++len) {
    int L = 1 << len;
    mp[L][L] = 1;
    for (int i = 1; i < L; ++i) {
      for (int j = 1; j < L; ++j) {
        mp[i + L][j + L] = mp[i + L][j] = mp[i][j + L] = mp[i][j];
      }
    }
  }
  if (swapped) swap(n, m);
  for (int i = 1; i < 1 << n; ++i) {
    for (int j = 1; j < 1 << m; ++j) {
      out << char('0' + (!swapped ? mp[i][j] : mp[j][i]));
    }
    out << '\n';
  }
}

