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>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef tuple<int,int,int> T;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
typedef vector<T> vt;
inline int getInt() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("Yes");}else{puts("No");}
const int MX = 200005;
typedef v(string) vs;



int main() {
  int n,m;
  scanf("%d%d",&n,&m);
  int h = 1<<n, w = 1<<m;
  vvi a(h,vi(w));
  rep(i,h)rep(j,w) a[i][j] = pcnt(i&j)%2;
  // rep(i,h) cout<<a[i]<<endl;
  vs ans(h-1, string(w-1,'0'));
  drep(i,h-1)drep(j,w) a[i+1][j] ^= a[i][j];
  drep(i,h)drep(j,w-1) a[i][j+1] ^= a[i][j];
  // rep(i,h)rep(j,i) {
  //   int cnt = 0;
  //   rep(k,w) cnt += a[i][k]^a[j][k];
  //   if (cnt != w/2) cout<<i<<" "<<j<<endl;
  // }
  rep(i,h-1)rep(j,w-1) {
    if (a[i+1][j+1]) ans[i][j] = '1';
  }
  rep(i,h-1) cout<<ans[i]<<endl;
  return 0;
}




















