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>
using namespace std;
const int N=2049;
int a[N][N],n,m;
void solve(int r1,int r2,int c1,int c2) {
	//printf("%d %d %d %d\n", r1,r2,c1,c2);
	if(r1==r2 && c1==c2) {
		a[r1][c1]=1;
		return;
	}
	int midr=(r1+r2)>>1, midc=(c1+c2)>>1;
	//printf("midr=%d,midc=%d\n", midr,midc);
	a[midr][midc]=1;
	solve(r1,midr-1,c1,midc-1);
	int d=(r2-r1)/2+1;
	for(int i=r1;i<=midr-1;i++){
		for(int j=c1;j<=midc-1;j++){
			a[i][j+d]=a[i][j];
			a[i+d][j]=a[i][j];
			a[i+d][j+d]=a[i][j];
		}
	}
}
int main() {
	scanf("%d%d",&n,&m);
	solve(0,(1<<10)-1,0,(1<<10)-1);
	for(int i=0;i<((1<<n)-1);i++){
		for(int j=0;j<((1<<m)-1);j++){
			printf("%d", a[i][j]);
		}
		printf("\n");
	}
}