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>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) int(a.size())

int gi() {
	int x=0,o=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m;
bool fl;

vector<vi> solve(int n) {
	//H W=2^n-1
	if(n==1) return {{1}};
	auto tmp=solve(n-1);
	int len=(1<<n)-1,len2=(1<<(n-1))-1;
	vector<vi> ret(len);
	for(int i=0;i<len;i++) ret[i].resize(len);
	for(int i=0;i<len2;i++)
		for(int j=0;j<len2;j++) ret[i][j]=ret[i+len2+1][j]=ret[i][j+len2+1]=ret[i+len2+1][j+len2+1]=tmp[i][j];
	ret[len2][len2]=1;return ret;
}

int main() {
	cin>>n>>m;
	if(n<m) fl=1,swap(n,m);
	auto ans=solve(n);
	if(!fl) {
		for(int i=0;i<(1<<n)-1;i++,cout<<'\n')
			for(int j=0;j<(1<<m)-1;j++) cout<<ans[i][j];
	}
	else {
		for(int i=0;i<(1<<m)-1;i++,cout<<'\n')
			for(int j=0;j<(1<<n)-1;j++) cout<<ans[j][i];
	}
	return 0;
}
