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 <map>
#include <set>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <fstream>
#include <bitset>
#include <queue>
#include <stack>
#include <deque>
#include <complex>
#include <iomanip>
#include <stdio.h>
#include <string.h>
#include <random>
#include <functional>

using std::cin;
using std::cout;
using std::cerr;
using std::endl;

using std::map;
using std::set;
using std::bitset;
using std::vector;
using std::string;
using std::multimap;
using std::multiset;
using std::deque;
using std::queue;
using std::stack;
using std::pair;
using std::iterator;

using std::sort;
using std::stable_sort;
using std::reverse;
using std::max_element;
using std::min_element;
using std::unique;
using std::ios_base;
using std::swap;
using std::fill;

using std::setprecision;
using std::fixed;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef string S;

ll min(ll a, ll b) {return a < b ? a : b;}
ll min(int a, ll b) {return a < b ? a : b;}
ll min(ll a, int b) {return a < b ? a : b;}
ll min(int a, int b) {return a < b ? a : b;}

ll max(ll a, ll b) {return a > b ? a : b;}
ll max(int a, ll b) {return a > b ? a : b;}
ll max(ll a, int b) {return a > b ? a : b;}
ll max(int a, int b) {return a > b ? a : b;}

namespace MySpace{

};

#define F(i, n) for (int (i) = 0; (i) != (n); (i)++)
#define fi first
#define se second
#define re return
#define all(x) (x).begin(), (x).end()
#define int long long

const int N = 5010;
const int MOD = 1e9 + 7;

int n, m;

int f(int x)
{
    int c = 0;
    while (x % 2 == 0) x /= 2, c++;
    return c;
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    n = (1 << n);
    m = (1 << m);
    for (int i = 1; i < n; cout << "\n", i++) for (int j = 1; j < m; j++) cout << (f(i) == f(j));
}
