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 <string>
#include <cstring>
#include <algorithm>
#define N 1234
template<typename T> inline void read(T &x) {
	x = 0; char c = getchar();
	while (!isdigit(c)) c = getchar();
	while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
}
using namespace std;
int h[N][N];
int n, m;
void Constr(int U, int D, int L, int R) {
	if (U == D && L == R)	return h[U][L] = 1, void();
	int m1 = (U + D)>>1, m2 = (L + R) >> 1;
	Constr(U, m1 - 1, L, m2 - 1);
	Constr(U, m1 - 1, m2 + 1, R);
	Constr(m1 + 1, D, L, m2 - 1);
	Constr(m1 + 1, D, m2 + 1, R);
	h[m1][m2] = 1;
}
int main() {
	read(n), read(m);
	n = (1 << n) - 1, m = (1 << m) - 1;
	int mx = max(n, m);
	Constr(1, mx, 1, mx);
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			printf("%d", h[i][j]);
		}
		putchar('\n');
	}
	return 0;
}
