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 <algorithm>
#include <utility>
#include <vector>
#include <numeric>

std::vector<std::vector<int>> generate(int rank) {
  if (rank == 1) {
    return { 
      { 0, 0 }, 
      { 0, 1 }
    };
  }
  int size = 1 << rank;
  int half = 1 << (rank - 1);
  std::vector<std::vector<int>> res(size, std::vector<int>(size));
  auto lower = generate(rank - 1);
  for (int i = 1; i < half; ++i) {
    for (int j = 1; j < half; ++j) {
      res[i][j] = lower[i][j];
      res[i + half - 1][j] = lower[i][j];
      res[i][size - j - 1] = lower[i][j];
      res[i + half - 1][size - j - 1] = lower[i][j] ^ 1;
    }    
  }
  for (int k = 0; k < half; ++k) {
    res[size - 1][size - k - 1] = 1;
    res[size - k - 1][size - 1] = 1;
  }
  return res;
}

int main() {
  int N, M;
  std::cin >> N >> M;
  auto base = generate(std::max(N, M));
  if (N > M) {
    for (int i = 0; i < (1 << N); ++i) {
      for (int j = 0; j < i; ++j) {
        std::swap(base[i][j], base[j][i]);
      }
    }
  }
  for (int i = 1; i < (1 << N); ++i) {
    for (int j = 1; j < (1 << M); ++j) {
      std::cout << (base[i - 1][j - 1] ^ base[i - 1][j] ^ base[i][j - 1] ^ base[i][j]);
    }
    std::cout << '\n';
  }
  return 0;
}
