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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define mem(x,v) memset(x,v,sizeof(x))
#define Rep(i,a,b) for(int i=(a);i<=int(b);++i)
#define rep(i,a,b) for(int i=(a);i<int(b);++i)
#define Dep(i,a,b) for(int i=(a);i>=int(b);--i)
#define gc getchar
#define pc putchar
#define fi first
#define se second
inline ll read(){
    ll x=0,f=1;char c=gc();
    for(;!isdigit(c);c=gc())if(c=='-')f=-1;
    for(;isdigit(c);c=gc())x=(x<<1)+(x<<3)+(c^48);
    return x*f;
}
inline void write(ll x){if(x<0)x=-x,pc('-');if(x>=10)write(x/10);putchar(x%10+'0');}
inline void writeln(ll x){write(x);pc('\n');}
inline void wri(ll x){write(x);pc(' ');}
int n,m,a[1<<11][1<<11];
#define lowbit(x) ((x)&-(x))
int main(){
	n = read(),m = read();
	int h = max(n,m);
	rep(i,1,1<<h)
		rep(j,1,1<<h)
			if(lowbit(i)==lowbit(j))
				a[i][j] = 1;
		rep(i,1,1<<n){
			rep(j,1,1<<m){
				pc(a[i][j]+'0');
			}puts("");
		}
	return 0;
}


