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>
#include <iomanip>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
#define ALL(a) a.begin(), a.end()
#define fi first
#define se second
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,K,Q,A,B;
string S;
const ll MOD = 998244353;
//const ll MOD = (1e+9) + 7;
const ll INF = 2e+9 + 10;
typedef pair<ll, ll> P;

mat make(int n){
    if(n == 1){
        return {{0, 0}, {0, 1}};
    }else{
        mat res = make(n - 1);
        rep(i,1<<(n - 1)) rep(j, 1<<(n - 1)) res[i].push_back(res[i][j]);
        rep(i, 1<<(n - 1)) res.push_back(res[i]);
        reps(i, 1<<(n - 1), 1<<n) reps(j, 1<<(n - 1), 1<<n) res[i][j] = res[i][j] ^ 1;
        return res;
    }
}

int main() {
    cin>>N>>M;
    mat memo = make(max(N, M));
    reps(i, 1, 1<<N) {
        reps(j, 1, 1 << M) cout << (memo[i][j] + memo[i][j - 1] + memo[i - 1][j] + memo[i - 1][j - 1])%2;
        cout << endl;
    }
}