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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

double startTime;
double getCurrentTime() {
	return ((double)clock() - startTime) / CLOCKS_PER_SEC;
}

const int N = 1035;
int a[N][N];
int b[N][N];

int main()
{
	startTime = (double)clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	for (int len = 1; 2 * len < N; len <<= 1) {
		for (int i = 0; i < len; i++)
			for (int j = 0; j < len; j++) {
				a[i][len + j] = a[len + i][len + j] = a[i][j];
				a[len + i][j] = a[i][j] ^ 1;
			}
	}
	
	int n, m;
	scanf("%d%d", &n, &m);
	n = 1 << n;
	m = 1 << m;
	for (int i = 0; i < n; i += min(n, m))
		for (int j = 0; j < m; j += min(n, m))
			for (int x = 0; x < min(n, m); x++)
				for (int y = 0; y < min(n, m); y++)
					b[i + x][j + y] = a[x][y];

	for (int i = 1; i < n; i++) {
		for (int j = 1; j < m; j++)
			printf("%d", a[i][j] ^ a[i - 1][j] ^ a[i][j - 1] ^ a[i - 1][j - 1]);
		printf("\n");
	}

	return 0;
}
