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 <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <cassert>

using namespace std;
typedef long long LL;
template <typename F, typename S> std::ostream& operator<<(ostream& os, const pair<F,S>& p) {
    os << "(" << p.first << "," << p.second << ")"; return os; }
template <typename T> ostream& operator<<(ostream& os, const vector<T>& v) {
    os << "["; for(const T& a: v){ os << a << ", "; } os << "]"; return os; }
template <typename K, typename V> ostream& operator<<(ostream& os, const map<K,V>& m) {
    os << "{"; for(const auto& p: m){ os << p.first <<":"<< p.second << ", "; } os << "}"; return os; }


int main(){
    LL n, m;
    cin >> n >> m;
    LL w = (1<<10);
    vector<vector<LL>> mat(w, vector<LL>(w));
    for(LL i = 0; i < 10; ++i){
        for(LL y = 0; y < (1<<i); ++y){
            for(LL x = 0; x < (1<<i); ++x){
                LL b = (1<<i);
                mat[b+y][x] = mat[y][x];
                mat[y][b+x] = mat[y][x];
                mat[b+y][b+x] = 1 - mat[y][x];
            }
        }
    }
    LL n2 = (1<<n)-1;
    LL m2 = (1<<m)-1;
    vector<vector<LL>> ans(n2, vector<LL>(m2));
    for(LL y = 0; y < n2; ++y){
        for(LL x = 0; x < m2; ++x){
            ans[y][x] = (mat[y+1][x+1] + mat[y][x+1] + mat[y+1][x] + mat[y][x]) % 2;
        }
    }
    for(LL y = 0; y < n2; ++y){
        for(LL x = 0; x < m2; ++x){
            cout << ans[y][x];
        }
        cout << endl;
    }
    return 0;
}

