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 rgi register int
#define ll long long
class fastin{
	private: int _ch,_f;
	public: template<typename _Tp> inline fastin& operator>>(_Tp&_x){
		_x=0; while(!isdigit(_ch)) _f|=(_ch==45),_ch=getchar();
		while(isdigit(_ch)) _x=(_x<<1)+(_x<<3)+(_ch^48),_ch=getchar();
		_f&&(_x=-_x,_f=0); return *this;}
	fastin() {_ch=_f=0;}
}fin;
class fastout{
	private: int _num[32],_head;
	public: int SUF; template<typename _Tp> inline fastout& operator<<(_Tp _x){
		_Tp _k; if(_x==0) {putchar(48),putchar(SUF);return *this;} if(_x<0) putchar(45),_x=-_x;
		while(_x>0) _k=_x/10,++_head,_num[_head]=(_x-(_k<<1)-(_k<<3))^48,_x=_k;
		while(_head>0) putchar(_num[_head]),--_head; putchar(SUF); return *this;}
	fastout() {_head=0,SUF=' ';}
}fout;
// ----------------------------
// #define int ll
// using namespace std;
const int maxn=2222;
const int mod=998244353,inf=1000000007;

bool fill[maxn][maxn];

void solve(int ax,int ay,int bx,int by)
{
	int xmid=(ax+bx)>>1,ymid=(ay+by)>>1;
	fill[xmid][ymid]=1;
	if(ax==bx&&ay==by) return;
	solve(ax,ay,xmid-1,ymid-1);
	solve(xmid+1,ay,bx,ymid-1);
	solve(ax,ymid+1,xmid-1,by);
	solve(xmid+1,ymid+1,bx,by);
}

signed main()
{
	int n,m,t;
	fin>>n>>m;
	t=std::max(n,m);
	solve(1,1,(1LL<<t)-1,(1LL<<t)-1);
	for(rgi i=1;i<1LL<<n;++i)
	{
		for(rgi j=1;j<1LL<<m;++j) printf("%d",fill[i][j]);
		printf("\n");
	}
	return 0;
}
// ----------------------------
// by imzzy
