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<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<list>
#include<bitset>
#include<stack>
#include<memory>
#include<numeric>
using namespace std;
using namespace std::chrono;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
#define REP(i, n) for(int i = 0;i < (n);i++)
/*cout<<fixed<<setprecision(20);cin.tie(0);ios::sync_with_stdio(false);*/
const int mod=998244353 ;
const llint inf=2.19e15+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-10;
template <class T,class U>bool chmin(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool chmax(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(const vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(const vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}

int main(void){
	static int ans[1024][1024]={};
	int h,i,j;
	for(h=0;h<=9;h++){
		int K=(1<<h);
		for(i=0;i<K;i++){
			for(j=0;j<K;j++){
				ans[i+K][j]=ans[i][j];
				ans[i][j+K]=ans[i][j];
				ans[i+K][j+K]=1-ans[i][j];
			}
		}
	}
	int n,m;cin>>n>>m;
	for(i=1;i<(1<<n);i++){
		for(j=1;j<(1<<m);j++){
			cout<<(ans[i][j]+ans[i-1][j]+ans[i][j-1]+ans[i-1][j-1])%2;
		}
		cout<<endl;
	}
}