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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se seoncd
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=1055;
int a[N][N];

void solve(int n,int m){
	if(n==1||m==1){
		rep(i,1,(1<<n)-1) rep(j,1,(1<<m))
			a[i][j]=1;
		return;
	}
	solve(n-1,m-1);
	int x=1<<(n-1),y=1<<(m-1);
	rep(i,1,x-1) rep(j,1,y-1)
		a[i+x][j]=a[i][j+y]
       =a[i+x][j+y]=a[i][j];
	rep(i,1,x-1) a[i][y]=a[i+x][y]=1;
	rep(i,1,y-1) a[x][i]=a[x][i+y]=1;
}

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