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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#include <complex>
#include <unordered_map>
#include <unordered_set>
#include <random>
#include <cassert>
#include <fstream>
#include <utility>
#include <functional>
#include <time.h>
#include <stack>
#include <array>
#define popcount __builtin_popcount
using namespace std;
typedef long long int ll;
typedef pair<ll, ll> P;
int s[1<<10][1<<10];
int ans[1<<10][1<<10];
int h[1<<10][1<<10];
void solve(int n){
    if(n==0) return;
    solve(n-1);
    for(int i=0; i<(1<<(n-1)); i++){
        for(int j=0; j<(1<<(n-1)); j++) h[i^(1<<(n-1))][j]=h[i][j], h[i][j^(1<<(n-1))]=h[i][j], h[i^(1<<(n-1))][j^(1<<(n-1))]=(1^h[i][j]);
    }
}
int main()
{
    int n, m;
    cin>>n>>m;
    bool rev=0;
    if(n>m) rev=1, swap(n, m);
    solve(n);
    for(int i=0; i<(1<<n); i++){
        for(int j=0; j<(1<<n); j++){
            if(h[i][j]){
                for(int k=0; k<(1<<(m-n)); k++) s[i][(j<<(m-n))^k]=1;
            }
        }
    }
    for(int i=1; i<(1<<n); i++){
        for(int j=1; j<(1<<m); j++){
            ans[i][j]=s[i][j-1]^s[i-1][j]^s[i-1][j-1]^s[i][j];
        }
    }
    if(rev){
        for(int i=1; i<(1<<m); i++){
            for(int j=1; j<(1<<n); j++) cout<<ans[j][i];
            cout<<endl;
        }
    }else{
        for(int i=1; i<(1<<n); i++){
            for(int j=1; j<(1<<m); j++) cout<<ans[i][j];
            cout<<endl;
        }
    }
    return 0;
}
