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
#pragma GCC optimize ("Ofast")
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i=(a); i<(int)(b); i++)
#define FORD(i, a, b) for (int i=a; i>(int)(b); i--)
#define PPC(x) __builtin_popcount(x)
#define MSB(x) (31 - __builtin_clz(x))
#define pb push_back
#define ALL(x) (x).begin(), (x).end()
#define ithBit(m, i) ((m) >> (i) & 1)
#define ft first
#define sd second
#ifdef DEBUG
#include "debug.h"
#else
#define dbg(...) 0
#endif
using namespace std;

const int maxN = 1 << 11;

int T[maxN][maxN];

void rob(int i, int j, int n)
{
	T[i+n/2][j+n/2] = 1;
	if (n == 1)
		return;
	rob(i, j, n/2);
	rob(i, j+n/2+1, n/2);
	rob(i+n/2+1, j, n/2);
	rob(i+n/2+1, j+n/2+1, n/2);
}

void solve()
{
	int n, m;
	scanf ("%d%d", &n, &m);
	n = (1<<n)-1, m = (1<<m)-1;
	
	rob(0, 0, max(n, m));
	
	FOR(i, 0, n)
	{
		FOR(j, 0, m)
			printf("%d", T[i][j]);
		printf("\n");
	}
} 

 
int main()
{
    int t;
    t=1;//scanf ("%d", &t);
    while (t--)
    	solve();
    return 0;    
}
