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 <string>
#include <vector>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
#include<functional>
#include<list>
#include<deque>
#include<bitset>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<cstring>
#include<sstream>
#include<complex>
#include<iomanip>
#include<numeric>
#include<cassert>
#define X first
#define Y second
#define pb push_back
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X))
#define rreps(X,S,Y) for (int (X) = (Y)-1;(X) >= (S);--(X))
#define repe(X,Y) for ((X) = 0;(X) < (Y);++(X))
#define peat(X,Y) for (;(X) < (Y);++(X))
#define all(X) (X).begin(),(X).end()
#define rall(X) (X).rbegin(),(X).rend()
#define eb emplace_back
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())
#define Endl endl
#define NL <<"\n"

using namespace std;
using ll=long long;
using pii=pair<int,int>;
using pll=pair<ll,ll>;
template<class T> using vv=vector<vector<T>>;
template<class T> inline bool MX(T &l,const T &r){return l<r?l=r,1:0;}
template<class T> inline bool MN(T &l,const T &r){return l>r?l=r,1:0;}
//#undef NUIP
#ifdef NUIP
#include "benri.h"
#else
#define out(args...)
#endif
#ifdef __cpp_init_captures
template<typename T>vector<T> table(int n, T v){ return vector<T>(n, v);}
template <class... Args> auto table(int n, Args... args){auto val = table(args...); return vector<decltype(val)>(n, move(val));}
#endif
const ll MOD=1e9+7; //998244353
vv<int> inv(vv<int> a){
	int n=a.size();
	int m=a[0].size();
	rreps(i,1,n)rep(j,m) a[i][j]^=a[i-1][j];
	rep(i,n)rreps(j,1,m) a[i][j]^=a[i][j-1];
	return a;
}
int oddness(vv<int> a){
	int n=a.size();
	int m=a[0].size();
	int re=0;
	rep(Y,n)rep(y,Y)rep(X,m)rep(x,X){
		if((a[y][x]+a[y][X]+a[Y][x]+a[Y][X])%2) ++re;
	}
	return re;
}

int bf(int n,int m){
	if(n*m>28) return 0;
	int mx=0;
	vv<int> re;
	rep(st,1<<(n*m)){
		vv<int> bd(n+1,vector<int>(m+1));
		rep(i,n)rep(j,m) bd[i+1][j+1]=st>>(i*m+j)&1;
		if(MX(mx,oddness(bd))) re=bd;
	}
	int cnt=0;
	rep(st,1<<(n*m)){
		vv<int> bd(n+1,vector<int>(m+1));
		rep(i,n)rep(j,m) bd[i+1][j+1]=st>>(i*m+j)&1;
		if(mx==oddness(bd)){
			++cnt;
			// auto re=bd;
			// auto re2=inv(re);
			// rep(i,n){
			// 	rep(j,m)cout<<".#"[re[i+1][j+1]];cout<<"\t";
			// 	rep(j,m)cout<<".#"[re2[i+1][j+1]];cout<<"\n";
			// } cout NL;
		}
	}
	out(n,m,mx,cnt,1);
	// auto re2=inv(re);
	// rep(i,n){
	// 	rep(j,m)cout<<".#"[re[i+1][j+1]];cout<<"\t";
	// 	rep(j,m)cout<<".#"[re2[i+1][j+1]];cout<<"\n";
	// 	} cout NL;
	return mx;
}

vv<int> solve(int n,int m){
	vv<int> bd(n+1,vector<int>(m+1));
	reps(i,1,n+1)reps(j,1,m+1){
		for(int d=1;d<112345;d*=2){
			if(i%d || j%d) break;
			bd[i][j]^=1;
		}
	}
	return bd;
}
vv<int> unko(int n,int m){
	vv<int> bd(n+1,vector<int>(m+1));
	reps(i,1,n+1)reps(j,1,m+1){
		vector<int> cnt(2);
		rep(ii,i)rep(jj,j) ++cnt[bd[ii][jj]^bd[ii][j]^bd[i][jj]];
		if(cnt[0]>cnt[1]) bd[i][j]=1;
	}
	return inv(bd);
}

int main(){
  ios_base::sync_with_stdio(false); cin.tie(0);
  cout<<fixed<<setprecision(0);
	// reps(i,1,6)reps(j,1,6){
	// 	int n=(1<<i)-1;
	// 	int m=(1<<j)-1;
	// 	auto re=solve(n,m);
	// 	auto er=unko(n,m);
	// 	if(re!=er){
	// 		rep(i,n+1){rep(j,m+1) cout<<".#"[re[i][j]^er[i][j]]; cout NL;}
	// 	}
	// }
	// return 0;
	int n,m;
	cin>>n>>m;
	n=(1<<n)-1;
	m=(1<<m)-1;
	auto re=solve(n,m);
	rep(i,n){rep(j,m) cout<<re[i+1][j+1]; cout NL;}
	return 0;
}
