Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

1 2

Sample Output 1

111

For this grid, S(1, 1, 1, 1), S(1, 1, 2, 2), S(1, 1, 3, 3), and S(1, 1, 1, 3) are odd, so it has the oddness of 4.

We cannot make the oddness 5 or higher, so this is one of the ways that maximize the oddness.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n)     for(int i = 0;i < n;i++)
#define REPR(i, n)    for(int i = n;i >= 0;i--)
#define FOR(i, m, n)  for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define VRSORT(v) sort(v.begin(), v.end(),greater<int>());
#define ALL(X) (X).begin(),(X).end()
#define INF 99999999
#define M_PI 3.14159265358979323846
#define MOD 1000000007

#ifdef __LOCAL
    #define DBG(X) cout << #X << " = " << (X) << endl;
    #define SAY(X) cout << (X) << endl;
#else
    #define DBG(X)
    #define SAY(X)
#endif

#ifdef __LOCAL
    #include <filesystem>
    namespace fs = std::filesystem;
#endif

using namespace std;

using ll = long long int;
using ull = unsigned long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

int dx[]={1, 0, -1, 0, 0};
int dy[]={0, 1, 0, -1, 0};
const int INT_INF = (int)(2e9);
const ll  LL_INF = (ll)(2e18);
const int NIL = -1;
static mt19937 _g(time(nullptr));
 
std::string pad(int num){ char buffer[4]; std::snprintf(buffer, sizeof(buffer), "%03d", num); return buffer;}
inline ll randint(ll a, ll b) { ll w = (_g() << 31LL) ^ _g(); return a + w % (b - a + 1); }
inline void fast_io() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); };
template<typename T> inline T sign(T x) { return T(x > 0) - T(x < 0); }
template<typename T, typename S> inline ostream& operator<<(ostream& os, const pair<T, S> p) { cout << "[" << p.first << ";" << p.second << "]"; return os; }
template<typename T> inline ostream& operator<<(ostream& os, const vector<T>& v) { for (auto el : v) os << el << " "; return os; }
template<typename T, typename S> inline ostream& operator<<(ostream& os, const map<T,S>& v) { for (auto el : v) os << el << " "; return os; }
template<typename T> inline T fetch() { T ret; cin >> ret; return ret; }
template<typename T> inline vector<T> fetch_vec(int sz) { vector<T> ret(sz); for (auto& elem : ret) cin >> elem; return ret; }

int N,M;
int row,column;
vector<vector<int>> A;

void input(){
  #ifdef __LOCAL
      fs::path p = __FILE__;
      fs::path input,output;
      input = output = p.parent_path();
      input += string("/input/") + string(p.stem()) + string(".txt");
      output += string("/output/") + string(p.stem()) + string(".txt");
      freopen(input.c_str(), "r", stdin);
      freopen(output.c_str(), "w", stdout);
  #endif

  fast_io();
  cin >> N >> M;
  row = pow(2,N)-1;
  column = pow(2,M)-1;
  A.assign(row,vector<int>(column,1));
}


int solve(){
  
  for (int i = 0; i < row; i++)
  {
    for (int j = 0; j < column; j++)
    {
      if(i%2==0)  cout << 1;
      else cout << (j%2==0);
    }
    cout << "\n";
  }
  
  return 0;
}

int main()
{
  input();
  solve();
  return 0;
}