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>
typedef long long int ll;
typedef long double ld;
#define pb push_back
#define pii pair < ll , ll >
#define F first
#define S second
#define endl '\n'
#define int long long
#define sync ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#define kill(x) return cout<<x<<'\n', 0;
using namespace std;
const int N=2000;
ll vis[N][N];
vector <vector <int> > f(ll x,ll y){
    memset(vis,0,sizeof vis);
    if (x==0){
       // cout << y << endl;
        vector <int> a;
        for (int i=0;i<=y;i++){
            a.pb((i+1)%2);
        }
        vector <vector <int> > z;
        z.pb(a);
        return z;
    }
    if (y==0){
        vector <vector <int> > y;
            vector <int> z;
            z.pb(1);
        for (int i=0;i<=x;i++){
            y.pb(z);
            z.clear();
            z.pb(i%2);
        }
        return y;
    }
    vector <vector <int> > z=f(x/2-1,y/2-1);
  //  cout << x/2 << " " << y/2 << " " << z.size() << " " << z[0][0] << endl;
    for (int i=0;i<=x;i++){
        for (int j=0;j<=y;j++){
            if (i==x/2 || j==y/2){
                vis[i][j]=1;
                if (i==x/2 && j==y/2) vis[i][j]=0;
                continue;
            }
            ll s=i,d=j;
            if (i>x/2) s-=(x+2)/2;
            if (j>y/2) d-=(y+2)/2;
            vis[i][j]=z[s][d];
        }
    }
    vector <vector <int> > ans;
    for (int i=0;i<=x;i++){
        vector <int> z;
        for (int j=0;j<=y;j++){
            z.pb(vis[i][j]);
        }
        ans.pb(z);
    }
    return ans;
}
int32_t main(){
    sync;
    ll n,m;
    cin >> n >> m;
    ll x=(1<<n)-1;
    ll y=(1<<m)-1;
    x--,y--;
    vector <vector <int> > a;
    a=f(x,y);
    for (int i=0;i<=x;i++){
        for (int j=0;j<=y;j++){
            cout << a[i][j] ;
        }
        cout << endl;
    }

}
