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
//Zory-2020
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
// typedef __int128 ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define FR first
#define SE second
#define MP make_pair
#define PB push_back
#define vc vector
#define db double
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bin(x) (1ll<<(x))
#define fo(i,l,r) for(int i=(l),I=(r);i<=I;i++)
#define fd(i,r,l) for(int i=(r),I=(l);i>=I;i--)
#define mem(x,val) memset(x,val,sizeof x)
#define Swap(a,b,n) for(int I=0;I<=n;I++) swap(a[I],b[I])
#define PC __builtin_popcountll
#ifdef DEBUG
	#define debug(A,args...) fprintf(stderr,A,##args)
#else
	#define debug(A,args...) printf("")
#endif
#define deb debug("line %d\n",__LINE__)
namespace mine
{
	ll qread()
	{
		ll ans=0,f=1;char c=getchar();
		while(c<'0' or c>'9') {if(c=='-')f=-1;c=getchar();}
		while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
		return ans*f;
	}
	void write(ll num){if(num<0) putchar('-'),num=-num;if(num>=10) write(num/10);putchar('0'+num%10);}
	void write1(ll num){write(num);putchar(' ');}
	void write2(ll num){write(num);putchar('\n');}
	template<typename T>inline bool chmax(T&a,const T&b){return a<b?a=b,1:0;}
	template<typename T>inline bool chmin(T&a,const T&b){return a>b?a=b,1:0;}
	ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
	bool IN(ll x,ll l,ll r){return l<=x and x<=r;}
	void GG(){puts("-1");exit(0);}
	
	const db eps=1e-8;
	const int INF=0x3f3f3f3f;
	const int MOD=1e9+7;
	int mm(const int x){return x>=MOD?x-MOD:x;}
	template<typename T> void add(T &x,const int &y){x=(x+y>=MOD?x+y-MOD:x+y);}
	ll qpower(ll x,ll e,int mod=MOD){ll ans=1;while(e){if(e&1)ans=ans*x%mod;x=x*x%mod;e>>=1;}return ans;}
	ll invm(ll x){return qpower(x,MOD-2);}
	const int M=1e6+10;
	ll fac[M],facinv[M],Inv[M];ll C(int n,int m){return n<0 or n<m?0:fac[n]*facinv[m]%MOD*facinv[n-m]%MOD;}
	void PRE()
	{
		fac[0]=1;fo(i,1,M-1) fac[i]=fac[i-1]*i%MOD;
		facinv[M-1]=invm(fac[M-1]);fd(i,M-1,1) facinv[i-1]=facinv[i]*i%MOD;
		Inv[1]=1;fo(i,2,M-1) Inv[i]=(MOD-MOD/i)*Inv[MOD%i]%MOD;
	}
	const int N=2e3+10;
	//------------------FIXED------------------
	int a[N][N],b[N];
	void solve(int n)
	{
		if(n==1){ a[1][1]=1;return; }
		solve(n-1);a[bin(n-1)][bin(n-1)]=1;
		fo(i,1,bin(n-1)-1) fo(j,1,bin(n-1)-1) a[i+bin(n-1)][j+bin(n-1)]=a[i+bin(n-1)][j]=a[i][j+bin(n-1)]=a[i][j];
	}
	void main()
	{
		int n=qread(),m=qread();bool tg=0;if(n<m) swap(n,m),tg=1;
		solve(n);
		if(!tg) fo(i,1,bin(n)-1) {fo(j,1,bin(m)-1) write(a[i][j]);puts("");}
		else fo(j,1,bin(m)-1) {fo(i,1,bin(n)-1) write(a[i][j]);puts("");}
	}
};//变量重名！
signed main()
{
	#ifdef DEBUG
	//freopen("a.in","r",stdin);
	freopen("z.txt","r",stdin);
	//freopen("a.out","w",stdout);
	#endif
	srand(time(0));
	mine::PRE();//线性预处理模意义
	mine::main();
	debug("\n---------------------Zory---------------------\nTime: %.2lf s",1.0*clock()/CLOCKS_PER_SEC);
}