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 double ld;
typedef long long ll;
typedef pair<double, double> pdd;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef vector<pii> vii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
//const int mod = ;

int main() {
  std::ios::sync_with_stdio(false); std::cin.tie(0);
  int n, m;
  cin >> n >> m;
  int tr = 0;
  if (n > m) {
    swap(n, m);
    tr = 1;
  }
  int N = (1 << n) - 1, M = (1 << m) - 1;
  vvi s(N, vi(M));
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      s[i][j] = __builtin_popcount((i + 1) & (j + 1)) % 2;
//      cerr << s[i][j];
    }
/*    cerr << endl;
    for (int i1 = 0; i1 < i; ++i1) {
      int c = 0;
      for (int j = 0; j < M; ++j) {
        c += s[i1][j] == s[i][j];
      }
      assert(c == M / 2);
    }*/
  }
  vvi res = s;
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < M; ++j) {
      if (i) res[i][j] ^= s[i - 1][j];
      if (j) res[i][j] ^= s[i][j - 1];
      if (i && j) res[i][j] ^= s[i - 1][j - 1];
    }
  }
  if (tr) {
    vvi res1(M, vi(N));
    for (int i = 0; i < res.size(); ++i) {
      for (int j = 0; j < res[0].size(); ++j) {
        res1[j][i] = res[i][j];
      }
    }
    res = res1;
  }
  for (int i = 0; i < res.size(); ++i) {
    for (int j = 0; j < res[0].size(); ++j) {
      cout << res[i][j];
    }
    cout << '\n';
  }
  return 0;
}
