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<iostream>
#include<vector>
#include<algorithm>
#define ep emplace
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef pair<int,int>pi;
typedef pair<ll,ll>pl;
const int inf=1e9+7;
const ll INF=1e18+7;
int arr[2050][2050];
void fill(int s,int e,int n,int m)
{
    if(n==0) {arr[s][e+(1<<m)-1]=1;return;}
    if(m==0) {arr[s+(1<<n)-1][e]=1;return;}
    arr[s+(1<<n)-1][e+(1<<m)-1]=1;
    fill(s,e,n-1,m-1);
    fill(s+(1<<n),e,n-1,m-1);
    fill(s,e+(1<<m),n-1,m-1);
    fill(s+(1<<n),e+(1<<m),n-1,m-1);
    return;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n,m;
    cin>>n>>m;
    fill(1,1,n-1,m-1);
    for(int i=1;i<1<<n;cout<<'\n',i++)for(int j=1;j<1<<m;j++)cout<<arr[i][j];
    cout.flush();
    return 0;
}