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,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define per(i,r,l) for (int i=r,i##end=l;i>=i##end;--i)
#define pb push_back
#define ll long long
using namespace std;
const int N=2003;
struct line{ int k,b; ll f(int x){ return (ll)k*x+b; } }a[N];
int A[N][N];
void solve(int n){
    if (n==2){ A[1][1]=1; return; }
    solve(n>>1); A[n>>1][n>>1]=1;
    rep(i,1,(n>>1)-1)
        rep(j,1,(n>>1)-1)
            A[i+(n>>1)][j]=A[i][j+(n>>1)]=A[i+(n>>1)][j+(n>>1)]=A[i][j];
}
int main(){
    int n,m; scanf("%d%d",&n,&m);
    int t=1<<max(n,m); solve(t);
    rep(i,1,(1<<n)-1){
        rep(j,1,(1<<m)-1) printf("%d",A[i][j]);
        putchar('\n');
    }
    return 0;
}
