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<string.h>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<vector>
#include<bitset>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<map>
using namespace std;
typedef long long ll;
typedef long double db;
typedef pair<int,int> pii;
const int N=100000;
const db pi=acos(-1.0);
#define lowbit(x) (x)&(-x)
#define sqr(x) (x)*(x)
#define rep(i,a,b) for (register int i=a;i<=b;i++)
#define per(i,a,b) for (register int i=a;i>=b;i--)
#define go(u,i) for (register int i=head[u];i;i=sq[i].nxt)
#define fir first
#define sec second
#define mp make_pair
#define pb push_back
#define maxd 998244353
#define eps 1e-8
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while ((ch<'0') || (ch>'9')) {if (ch=='-') f=-1;ch=getchar();}
    while ((ch>='0') && (ch<='9')) {x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

namespace My_Math{
	#define N 100000

	int fac[N+100],invfac[N+100];

	int add(int x,int y) {return x+y>=maxd?x+y-maxd:x+y;}
	int dec(int x,int y) {return x<y?x-y+maxd:x-y;}
	int mul(int x,int y) {return 1ll*x*y%maxd;}
	ll qpow(ll x,int y)
	{
		ll ans=1;
		while (y)
		{
			if (y&1) ans=mul(ans,x);
			x=mul(x,x);y>>=1;
		}
		return ans;
	}
	int inv(int x) {return qpow(x,maxd-2);}

	int C(int n,int m)
	{
		if ((n<m) || (n<0) || (m<0)) return 0;
		return mul(mul(fac[n],invfac[m]),invfac[n-m]);
	}

	int math_init()
	{
		fac[0]=invfac[0]=1;
		rep(i,1,N) fac[i]=mul(fac[i-1],i);
		invfac[N]=inv(fac[N]);
		per(i,N-1,1) invfac[i]=mul(invfac[i+1],i+1);
	}
	#undef N
}
using namespace My_Math;

int n,m,a[2020][2020];

void solve(int l1,int r1,int l2,int r2)
{
	if ((l1==l2) && (r1==r2)) {a[l1][r1]=1;return;}
	int midl=(l1+l2)>>1,midr=(r1+r2)>>1;
	a[midl][midr]=1;
	solve(l1,r1,midl-1,midr-1);
	solve(l1,midr+1,midl-1,r2);
	solve(midl+1,r1,l2,midr-1);
	solve(midl+1,midr+1,l2,r2);
}

int main()
{
	n=read();m=read();
	int mx=max(n,m);
	solve(1,1,(1<<mx)-1,(1<<mx)-1);
	rep(i,1,(1<<n)-1) 
	{
		rep(j,1,(1<<m)-1) putchar('0'+a[i][j]);
		puts("");
	}
	return 0;
}

