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>
using namespace std;
using ll=long long;
#define fr(i,n) for(int i=0;i<(n);++i)
#define Fr(i,n) for(int i=1;i<=(n);++i)
#define ifr(i,n) for(int i=(n)-1;i>=0;--i)
#define iFr(i,n) for(int i=(n);i>0;--i)

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    istream& in(cin);
    ostream& out(cout);
    bool f=false,ans[1024][1024]={};
    int n,m;
    in>>n>>m;
    if(n<m){
        swap(n,m);
        f=true;
    }
    int t=1;
    fr(_,m){
        fr(i,t) fr(j,t){
            ans[i+t][j]=ans[i][j+t]=ans[i][j];
            ans[i+t][j+t]=!ans[i][j];
        }
        t<<=1;
    }
    int u=t;
    fr(_,n-m){
        fr(i,u) fr(j,t){
            ans[i+u][j]=ans[i][j];
        }
        u<<=1;
    }
    /*if(f){
        fr(i,(1<<m)){
            fr(j,(1<<n)) cout<<ans[j][i];
            cout<<endl;
        }
    }
    else{
        fr(i,(1<<n)){
            fr(j,(1<<m)) cout<<ans[i][j];
            cout<<endl;
        }
    }
    cout<<endl;*/
    fr(i,1<<n){
        iFr(j,(1<<m)-1){
            //cout<<i<<" "<<j<<" "<<ans[i][j]<<" "<<ans[i][j-1]<<" ";
            ans[i][j]=ans[i][j]^ans[i][j-1];
            //cout<<ans[i][j]<<endl;
        }
    }
    /*if(f){
        fr(i,(1<<m)){
            fr(j,(1<<n)) cout<<ans[j][i];
            cout<<endl;
        }
    }
    else{
        fr(i,(1<<n)){
            fr(j,(1<<m)) cout<<ans[i][j];
            cout<<endl;
        }
    }
    cout<<endl;*/

    iFr(i,(1<<n)-1){
        fr(j,1<<m){
            //cout<<i<<" "<<j<<" "<<ans[i][j]<<" "<<ans[i-1][j]<<" ";
            ans[i][j]=ans[i][j]^ans[i-1][j];
            //cout<<ans[i][j]<<endl;
        }
    }
    if(f){
        Fr(i,(1<<m)-1){
            Fr(j,(1<<n)-1) cout<<ans[j][i];
            cout<<endl;
        }
    }
    else{
        Fr(i,(1<<n)-1){
            Fr(j,(1<<m)-1) cout<<ans[i][j];
            cout<<endl;
        }
    }
}