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
/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author aajisaka
 */

#include<bits/stdc++.h>

using namespace std;

void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#define SPEED ios_base::sync_with_stdio(false);cin.tie(nullptr)
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define all(v) v.begin(), v.end()
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

using ll = long long;
using P = pair<ll, ll>;

constexpr long double PI = 3.14159265358979323846264338327950288L;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

class EOddSumRectangles {
public:
    void solve(istream& cin, ostream& cout) {
      SPEED;
      int n, m; cin >> n >> m;
      int ma = max(n, m);
      int h = (1<<ma);
      int w = (1<<ma);
      int dp[h][w];
      dp[0][0] = dp[0][1] = dp[1][0] = 0;
      dp[1][1] = 1;

      for(int k=2; k<=ma; k++) {
        int r = 1<<(k-1);
        rep(i, 1<<k) {
          rep(j, 1<<k) {
            if (i < r && j < r) continue;
            if (i < r) {
              dp[i][j] = dp[i][j-r];
            } else if (j < r) {
              dp[i][j] = dp[i-r][j];
            } else {
              dp[i][j] = 1-dp[i-r][j-r];
            }
          }
        }
      }

      rep(i, 1<<ma) {
        rep(j, 1<<ma) {
          cerr << dp[i][j];
        }
        cerr << endl;
      }

      rep(i, (1<<n)-1) {
        rep(j, (1<<m)-1) {
          if (dp[i+1][j] ^ dp[i][j+1] ^ dp[i][j]) {
            cout << 1 - dp[i+1][j+1];
          } else {
            cout << dp[i+1][j+1];
          }
        }
        cout << '\n';
      }
    }
};

signed main() {
  EOddSumRectangles solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}