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>
#define DEBUG fprintf(stderr, "Passing [%s] line %d\n", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

inline int gi()
{
	int f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

inline LL gl()
{
	LL f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 1030, M = N << 1;

int n, m, b[N][N], a[N][N];

inline int popcnt(int x)
{
	int res = 0;
	while (x)
	{
		if (x & 1) ++res;
		x >>= 1;
	}
	return res;
}

int main()
{
	//File("");
	n = gi(), m = gi();
	int r = 1, c = 1;
	for (int i = 1; i <= n; i+=1) r <<= 1;
	for (int i = 1; i <= m; i+=1) c <<= 1;
	--r, --c;
	for (int i = 1; i <= r; i+=1)
		for (int j = 1; j <= c; j+=1)
			b[i][j] = popcnt(i & j) % 2;
	for (int i = 1; i <= r; i+=1)
	{
		for (int j = 1; j <= c; j+=1)
			a[i][j] = ((b[i][j] - b[i - 1][j] - b[i][j - 1] + b[i - 1][j - 1]) % 2 + 2) % 2,
				printf("%d", a[i][j]);
		puts("");
	}
	return 0;
}
