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;

int main(){
	int n,m;
	cin>>n>>m;
	
	if(n<=m){
		string s[12];
		vector<int> t;
		t.push_back((1<<(m-1))-1);
		for(int i=0;i<m;i++){
			for(int j=0;j<(1<<m)-1;j++){
				s[i]+='0';
			}
			vector<int> t_;
			for(int x: t){
				s[i][x]='1';
				if(i+1<m){
					t_.push_back(x-(1<<(m-2-i)));
					t_.push_back(x+(1<<(m-2-i)));
				}
			}
			t=t_;
		}
		for(int i=1;i<=(1<<n)-1;i++){
			int t=0;
			while(((i>>t)&1)==0)t++;
			cout<<s[t]<<endl;
		}
	}
	else {
		swap(n,m);
		string s[12];
		vector<int> t;
		t.push_back((1<<(m-1))-1);
		for(int i=0;i<m;i++){
			for(int j=0;j<(1<<m)-1;j++){
				s[i]+='0';
			}
			vector<int> t_;
			for(int x: t){
				s[i][x]='1';
				if(i+1<m){
					t_.push_back(x-(1<<(m-2-i)));
					t_.push_back(x+(1<<(m-2-i)));
				}
			}
			t=t_;
		}
		char ret[2000][2000];
		for(int i=1;i<=(1<<n)-1;i++){
			int t=0;
			while(((i>>t)&1)==0)t++;
			for(int j=1;j<=s[t].size();j++){
				ret[i][j]=s[t][j-1];
			}
		}
		for(int i=1;i<=(1<<m)-1;i++){
			for(int j=1;j<=(1<<n)-1;j++){
				printf("%c",ret[j][i]);
			}
			printf("\n");
		}
	}
}

