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 long ll;
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)
#define rep2(i, a, n) for(ll i = a; i < (ll)(n); i++)
#define memi cout << endl
#define kono(n) cout << fixed << setprecision(n)
#define all(c) (c).begin(), (c).end()
#define pb push_back
#define hina cout << ' '
#define in(n) cin >> n
#define in2(n, m) cin >> n >> m
#define in3(n, m, l) cin >> n >> m >> l
#define out(n) cout << n
const ll mei = (ll)1e9 + 7;

int main(){
  ll a, b, c, d, i, j, m, n;
  in2(a, b);
  c = d = 1;
  rep(k, a)
    c = c * 2;
  rep(k, b)
    d = d * 2;
  a = c / 2;
  b = d / 2;
  m = c;
  n = d;
  vector<vector<ll>> g(c, vector<ll> (d, 1));
  while(true){
    i = 0;
    if(a <= 1 || b <= 1)
      break;
    else{
      while(a + m * i < c){
        j = 0;
        while(b + n * j < d){
          g[a + m * i - 1][b + n * j - 1] = 0;
          j++;
        }
        i++;
      }
      if(a > 1){
        a = a / 2;
        m = m / 2;
      }
      if(b > 1){
        b = b / 2;
        n = n / 2;
      }
    }
  }
  rep(k, c - 1){
    rep(l, d - 1)
      out(g[k][l]);
    memi;
  }
}