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 f first
#define s second
#define MOD 1000000007
#define PMOD 998244353
#define pb(x) push_back(x)
using namespace std;

typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> plii;
typedef pair<int, pii> piii;
const int INF = 1e9+10;
const ll LINF = 1LL*INF*INF;
const int MAXN = 2e5+10;
const int MAXM = 5e3+10;

priority_queue<int> pq;
vector<vector<int> > graph;
queue<int> que;

int A[MAXN];
char S[MAXN];

int main()
{
    int n,m,k,a,b,x,y,q;
    int sum = 0;
    int cnt = 0;
    int mx = 0;
    int mn = INF;
    int cur = 0, idx = -1;
    int tc;

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin>>n>>m;

    for(int i=1;i<(1<<n);i++){
        for(int j=1;j<(1<<m);j++){
            if((i&-i)==(j&-j))S[j] = '1';
            else S[j] = '0';
        }
        cout<<S+1<<"\n";
    }

    return 0;
}
