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 <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <iomanip>
#include <limits>
#include <list>
#include <queue>
#include <tuple>
#include <map>
#include <stack>
#include <set>
#include <bitset>
#include <functional>
using namespace std;
#define fast_io ios_base::sync_with_stdio (false) ; cin.tie(0) ; cout.tie(0) ;
#define ll long long int
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define reps(i,n) for(int i=1; i<=(int)(n); i++)
#define REP(i,n) for(int i=n-1; i>=0; i--)
#define REPS(i,n) for(int i=n; i>0; i--)
#define MOD (long long int)(1e9+7)
#define INF (int)(1123456789)
#define LINF (long long int)(112345678901234567)
#define chmax(a, b) a = (((a)<(b)) ? (b) : (a))
#define chmin(a, b) a = (((a)>(b)) ? (b) : (a))
#define all(v) v.begin(), v.end()
typedef pair<int, int> Pii;
typedef pair<ll, ll> Pll;

ll mpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = mpow(a,b/2); return memo*memo%MOD;}
  else return mpow(a,b-1) * a % MOD;
}
ll lpow(ll a, ll b){
  if(b==0) return 1;
  else if(b%2==0){ll memo = lpow(a,b/2); return memo*memo;}
  else return lpow(a,b-1) * a;
}
ll gcd(ll a, ll b){
  if(b==0) return a;
  else return gcd(b, a%b);
}
vector<ll> kaijo_memo;
ll kaijo(ll n){
  if(kaijo_memo.size() > n) return kaijo_memo[n];
  if(kaijo_memo.size() == 0) kaijo_memo.push_back(1);
  while(kaijo_memo.size() <= n) kaijo_memo.push_back(kaijo_memo[kaijo_memo.size()-1] * kaijo_memo.size() % MOD);
  return kaijo_memo[n];
}
vector<ll> gyaku_kaijo_memo;
ll gyaku_kaijo(ll n){
  if(gyaku_kaijo_memo.size() > n) return gyaku_kaijo_memo[n];
  if(gyaku_kaijo_memo.size() == 0) gyaku_kaijo_memo.push_back(1);
  while(gyaku_kaijo_memo.size() <= n) gyaku_kaijo_memo.push_back(gyaku_kaijo_memo[gyaku_kaijo_memo.size()-1] * mpow(gyaku_kaijo_memo.size(), MOD-2) % MOD);
  return gyaku_kaijo_memo[n];
}

ll nCr(ll n, ll r){
  if(n == r) return 1;//0個の丸と-1個の棒みたいな時に時に効く？不安.
  if(n < r || r < 0) return 0;
  ll ret = 1;
  if(n <= 1e7){
    ret *= kaijo(n); ret %= MOD;
    ret *= gyaku_kaijo(r); ret %= MOD;
    ret *= gyaku_kaijo(n-r); ret %= MOD;
  }else{
    rep(i,r){
      ret *= n-i; ret %= MOD;
      ret *= mpow(r-i, MOD-2); ret %= MOD;
    }
  }
  return ret;
}

int main(void){
  fast_io
  cout<<fixed<<setprecision(15);

  int n,m;cin>>n>>m;
  bool hoge = false;
  if(n < m){
    swap(n,m);
    hoge = true;
  }

  vector<vector<int>> S(lpow(2,m));
  rep(j,lpow(2,n)){
    S[0].push_back(0);
  }
  ll k = 1;
  reps(i,lpow(2,m)-1){
    if(k*2 == i){
      k *= 2;
    }
    rep(j,lpow(2,n)){
      if(j/k%2 == 0){
        S[i].push_back(S[i-k][j]);
      }else{
        S[i].push_back((S[i-k][j] + 1)%2);
      }
    }
  }
/*
  rep(i,lpow(2,m)){
    rep(j,lpow(2,n)){
      cout<<S[i][j];
    }
    cout<<endl;
  }
  cout<<endl;
*/

  rep(i,lpow(2,m)){
    REP(j,lpow(2,n)-1){
      S[i][j+1] = (S[i][j+1] + S[i][j]) % 2;
    }
  }
/*
  reps(i,lpow(2,m)-1){
    reps(j,lpow(2,n)-1){
      cout<<S[i][j];
    }
    cout<<endl;
  }
  cout<<endl;
*/
  rep(j,lpow(2,n)){
    REP(i,lpow(2,m)-1){
      S[i+1][j] = (S[i+1][j] + S[i][j]) % 2;
    }
  }
  if(hoge){
  reps(i,lpow(2,m)-1){
    reps(j,lpow(2,n)-1){
      cout<<S[i][j];
    }
    cout<<endl;
  }}else{

  reps(j,lpow(2,n)-1){
    reps(i,lpow(2,m)-1){
      cout<<S[i][j];
    }
    cout<<endl;
  }}
}