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<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#include<bitset>
#include<set>
#include<assert.h>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define pll pair<long long,long long>
#define pii pair<int,int>
#define SZ(x) ((int)x.size())
using namespace std;
inline LL read()
{
	LL f=1,x=0;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;
}
int stack[20];
template<typename T>inline void write(T x)
{
	if(x<0){putchar('-');x=-x;}
    if(!x){putchar('0');return;}
    int top=0;
    while(x)stack[++top]=x%10,x/=10;
    while(top)putchar(stack[top--]+'0');
}
template<typename T>inline void pr1(T x){write(x);putchar(' ');}
template<typename T>inline void pr2(T x){write(x);putchar('\n');}
template<typename T>inline void chkmin(T &x,T y){x=x<y?x:y;}
template<typename T>inline void chkmax(T &x,T y){x=x>y?x:y;}
const int MAXN=1030;
int a[MAXN][MAXN],n,m;
int main()
{
	n=read();m=read();int h=max(n,m);
	a[1][1]=1;
	for(int i=2;i<=h;i++)
	{
		int lm=(1<<(i-1));
		for(int j=lm;j<2*lm;j++)for(int k=0;k<lm;k++)
			a[j][k]=a[j-lm][k];
		for(int j=0;j<lm;j++)for(int k=lm;k<2*lm;k++)
			a[j][k]=a[j][k-lm];
		for(int j=lm;j<2*lm;j++)for(int k=lm;k<2*lm;k++)
			a[j][k]=a[j-lm][k-lm]^1;
	}
	if(n<=m)
	{
		for(int i=1;i<(1<<n);i++)
		{
			for(int j=1;j<(1<<m);j++)write(a[i][j]^a[i-1][j]^a[i][j-1]^a[i-1][j-1]);
			puts("");
		}
	}
	else
	{
		for(int j=1;j<(1<<n);j++)
		{
			for(int i=1;i<(1<<m);i++)write(a[i][j]^a[i-1][j]^a[i][j-1]^a[i-1][j-1]);
			puts("");
		}
	}
	return 0;
}
