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>

#if MYDEBUG
#include "lib/cp_debug.hpp"
#else
#define DBG(...) ;
#endif
#if __cplusplus <= 201402L
template <typename T>
T gcd(T a, T b) { return ((a % b == 0) ? b : gcd(b, a % b)); }
template <typename T>
T lcm(T a, T b) { return a / gcd(a, b) * b; }
#endif
using LL = long long;
constexpr LL LINF = 334ll << 53;
constexpr int INF = 15 << 26;
constexpr LL MOD = 1E9 + 7;

namespace Problem {
using namespace std;

class Helper {
 public:
  int n, m;
  vector<vector<int>> a;
  Helper(LL n, LL m) : n(n), m(m), a(n, vector<int>(m)){};

  void solve() {
    int N = (1 << n) - 1;
    int M = (1 << m) - 1;
    int ma = 0;
    vector<vector<int>> ans;
    a = vector<vector<int>>(N, vector<int>(M));
    vector<vector<int>> b(N + 1, vector<int>(M + 1));
    DBG(N, M)
    for (LL mask = 0; mask < (1ll << (N * M)); ++mask) {
      for (int i = 0; i < N; ++i) {
        for (int j = 0; j < M; ++j) {
          int p = i * M + j;
          a[i][j] = (mask >> p) & 1;
          b[i + 1][j + 1] = a[i][j];
        }
      }
      for (int i = 1; i <= N; ++i) {
        for (int j = 1; j <= M; ++j) {
          b[i][j] += b[i - 1][j];
        }
      }

      for (int i = 1; i <= N; ++i) {
        for (int j = 1; j <= M; ++j) {
          b[i][j] += b[i][j - 1];
        }
      }
      int cnt = 0;
      for (int i1 = 1; i1 <= N; ++i1) {
        for (int i2 = i1; i2 <= N; ++i2) {
          for (int j1 = 1; j1 <= M; ++j1) {
            for (int j2 = j1; j2 <= M; ++j2) {
              int sum = b[i2][j2] - b[i1 - 1][j2] - b[i2][j1 - 1] + b[i1 - 1][j1 - 1];
              cnt += sum % 2;
            }
          }
        }
      }
      //DBG(a, cnt)
      if (cnt > ma) {
        ans = a;
        ma = cnt;
      }
    }
    DBG(ans)
    DBG(ma)
  }
};
class Solver2 {
 public:
  int n, m;
  Solver2(LL n, LL m) : n(n), m(m){};

  void solve() {
    int N = (1 << n) - 1;
    int M = (1 << m) - 1;
    auto a = vector<vector<int>>(N, vector<int>(M));
    DBG(N, M)
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < M; ++j) {
        int cntn = 0;
        int cntm = 0;
        int ti = i + 1, tj = j + 1;
        while (ti % 2 == 0) {
          cntn++;
          ti /= 2;
        }
        while (tj % 2 == 0) {
          cntm++;
          tj /= 2;
        }
        if (cntn >= cntm)
          a[i][j] = 1;
      }
    }
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j < M; ++j) {
        cout << a[i][j];
      }
      cout << endl;
    }
  }
};
}  // namespace Problem

int main() {
  std::cin.tie(0);
  std::ios_base::sync_with_stdio(false);
  // std::cout << std::fixed << std::setprecision(12);
  long long n = 0, m;
  std::cin >> n >> m;

  Problem::Solver2 sol(n, m);
  sol.solve();
  return 0;
}
