#include <bits/stdc++.h>  // clang-format off
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define each(x,v) for(auto& x : v)
#define all(v) (v).begin(),(v).end()
#define sz(v) ((int)(v).size())
#define ini(...) int __VA_ARGS__; in(__VA_ARGS__)
#define inl(...) long long __VA_ARGS__; in(__VA_ARGS__)
#define ins(...) string __VA_ARGS__; in(__VA_ARGS__)
#define inc(...) char __VA_ARGS__; in(__VA_ARGS__)
#define in2(s,t)     rep(i,sz(s)){in(s[i] , t[i]);}
#define in3(s,t,u)   rep(i,sz(s)){in(s[i] , t[i] , u[i]);}
#define in4(s,t,u,v) rep(i,sz(s)){in(s[i] , t[i] , u[i] , v[i]);}

#ifdef ONLINE_JUDGE
  #define rep(i,N) for(int i = 0; i < (int)(N); i++)
  #define repr(i,N) for(int i = (int)(N) - 1; i >= 0; i--)
  #define rep1(i,N) for(int i = 1; i <= (int)(N) ; i++)
  #define repr1(i,N) for(int i = (N) ; (int)(i) > 0 ; i--)
#else
  #define rep(i,N) for(long long i = 0; i < (long long)(N); i++)
  #define repr(i,N) for(long long i = (long long)(N) - 1; i >= 0; i--)
  #define rep1(i,N) for(long long i = 1; i <= (long long)(N) ; i++)
  #define repr1(i,N) for(long long i = (N) ; (long long)(i) > 0 ; i--)
#endif

using namespace std; void solve();
using ll = long long; template<class T = ll> using V = vector<T>;
using vi = V<int>; using vl = V<>; using vvi = V< V<int> >;
using vd = V<double>; using vs = V<string>; using vvl = V< V<> >;
constexpr int inf = 1001001001; constexpr ll infLL = (1LL << 61) - 1;
struct IoSetupNya {IoSetupNya() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); cerr << fixed << setprecision(7);} } iosetupnya;
template<typename T, typename U> inline bool amin(T &x, U y) { return (y < x) ? (x = y, true) : false; }
template<typename T, typename U> inline bool amax(T &x, U y) { return (x < y) ? (x = y, true) : false; }
template<typename T,typename U>ll ceil(T a,U b){return (a + b - 1) / b;}
template<typename T, typename U> ostream& operator <<(ostream& os, const pair<T, U> &p) { os << p.first << " " << p.second; return os; }
template<typename T, typename U> istream& operator >>(istream& is, pair<T, U> &p) { is >> p.first >> p.second; return is; }
template<typename T> ostream& operator <<(ostream& os, const vector<T> &v) { int s = (int)v.size(); for(int i=0;i<s;i++) os << (i ? " " : "") << v[i]; return os; }
template<typename T> istream& operator >>(istream& is, vector<T> &v) { for(auto &x : v) is >> x; return is; }
void in(){} template <typename T,class... U> void in(T &t,U &...u){ cin >> t; in(u...);}
void out(){cout << "\n";} template <typename T,class... U> void out(const T &t,const U &...u){ cout << t; if(sizeof...(u)) cout << " "; out(u...);}
template<typename T>void die(T x){out(x); exit(0);}

#ifdef NyaanDebug
  #include "NyaanDebug.h"
  #define trc(...) do { cerr << #__VA_ARGS__ << " = "; dbg_out(__VA_ARGS__);} while(0)
  #define trca(v,N) do { cerr << #v << " = "; array_out(v , N);} while(0)
  #define trcc(v) do { cerr << "name : " << #v << "\n"; int cnt = 0; each(x , v){cerr << (cnt++) << " : "; trc(x); } } while(0)
#else
  #define trc(...)
  #define trca(...)
  #define trcc(...)
  int main(){solve();}
#endif

using P = pair<ll,ll>; using vp = V<P>;
constexpr int MOD = /** 1000000007; //*/ 998244353;
// clang-format on
/////////////////////////

void solve(){
  
  ini(N , M);
  int H = (1 << N) - 1;
  int W = (1 << M) - 1;
  vvi ans(1023 , vi(1023 , 1));

  auto f = [&](auto rec,int l,int u,int s) -> void{
    if(s == 3){
      ans[l + 1][u + 1] = 0;
    }
    else{
      int d = (s + 1) / 2;
      ans[l + d - 1][u + d - 1] = 0;
      rec(rec , l , u , s / 2);
      rec(rec , l + d, u + d, s / 2);
      rec(rec , l + d, u , s / 2);
      rec(rec , l , u + d, s / 2);
    }
  };

  f(f , 0 , 0 , 1023);

  rep(i,H){
    rep(j,W)cout<< ans[i][j];
    cout << '\n';
  }
  

 /*
  int N = 3 , M = 7;
  vvi memo( N, vi(M));
  vvi ans;
  int score = 0;
  auto f = [&](int a,int b,int c,int d){
    int ret = 0;
    for(int i = a ; i <= b ; i++)
      for(int j = c ; j <= d ; j++)
        ret ^= memo[i][j];
    return ret;
  };
  rep(b , 1LL << (N * M)){
    rep(i,N) rep(j,M){
      memo[i][j] = (b >> (i * M + j)) & 1;
    }
    int cur = 0;
    rep(b,N) rep(a,b+1) rep(d,M) rep(c,d+1){
      //trc(a,b,c,d);
      cur += f(a,b,c,d);
    }
    if(amax(score , cur)) ans = memo;
    if(cur == 96){
      each(x,memo) out(x);
      out();
    }
  }
  out(score);
  out(ans);
  */
  

}

Measured memory_bytes: 5376