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=(b)-1;i>=(a);--i)
#define ll long long
#define lb(x) ((x)&(-x))
#define pii pair<int,int>
#define vi vector<int>
#define pb push_back
#define fi first
#define se second
#define de(x) cout<<x<<" x"<<endl
#define LCAFA rep(i,1,20)rep(j,1,n+1)fa[j][i]=fa[fa[j][i-1]][i-1]
#define all(x) x.begin(),x.end()
#define ls(x) x<<1
#define rs(x) x<<1|1
using namespace std;
const int N=2e6+9;
const ll mod=998244353;
char s[1050];
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	rep(i,1,1<<n){
		rep(j,1,1<<m){
			if(lb(i)!=lb(j))s[j]='0';
			else s[j]='1';
		}
		printf("%s\n",s+1);
	}
} 