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;

int calc(vector<vector<int>> &a){
	int i1, i2, j1, j2, k, l, ans = 0;
	for(i1 = 0; i1 < a.size(); i1++){
		for(i2 = i1; i2 < a.size(); i2++){
			for(j1 = 0; j1 < a[0].size(); j1++){
				for(j2 = j1; j2 < a[0].size(); j2++){
					int now = 0;
					for(k = i1; k <= i2; k++){
						for(l = j1; l <= j2; l++){
							now += a[k][l];
						}
					}
					ans += now % 2;
				}
			}
		}
	}
	return ans;
}

vector<vector<int>> make(int A, int B){
	int i, j;
	vector<vector<int>> ans(A, vector<int>(B, -1));
/*	if(A == 1){
		for(j = 0; j < B; j++){
			ans[0][j] = 1 - j % 2;
		}
		return ans;
	}
	if(B == 1){
		for(i = 0; i < A; i++){
			ans[i][0] = 1 - i % 2;
		}
		return ans;
	}
*/	if(A * B == 0){
		return ans;
	}
	vector<vector<int>> res = make(A / 2, B / 2);
/*	for(i = 0; i < A / 2; i++){
		for(j = 0; j < B / 2; j++){
			printf("%d", res[i][j]);
		}
		printf("\n");
	}
	printf("\n");
*/	for(i = 0; i < A; i++){
		ans[i][B / 2] = 0;
	}
	for(j = 0; j < B; j++){
		ans[A / 2][j] = 0;
	}
	ans[A / 2][B / 2] = 1;
/*	for(i = 0; i < A; i++){
		for(j = 0; j < B; j++){
			printf("%d", ans[i][j]);
		}
		printf("\n");
	}
	printf("\n");
*/	for(i = 0; i < A / 2; i++){
		for(j = 0; j < B / 2; j++){
			ans[i][j] = res[i][j];
			ans[A / 2 + 1 + i][j] = res[i][j];
			ans[i][B / 2 + 1 + j] = res[i][j];
			ans[A / 2 + 1 + i][B / 2 + 1 + j] = res[i][j];
		}
	}
/*	for(i = 0; i < A; i++){
		for(j = 0; j < B; j++){
			printf("%d", ans[i][j]);
		}
		printf("\n");
	}
	printf("\n");
*/	return ans;
}

int main(){
	int N, M, i, j;
	scanf("%d%d", &N, &M);
	vector<vector<int>> a((1 << N) - 1, vector<int>((1 << M) - 1, -1));
/*	for(i = 0; i < (1 << N) - 1; i++){
		for(j = 0; j < (1 << M) - 1; j++){
			a[i][j] = 1 - (i + j) % 2;
			scanf("%1d", &a[i][j]);
		}
	}
*/	a = make((1 << N) - 1, (1 << M) - 1);
//	printf("score = %d\n", calc(a));
//	return 0;
	for(i = 0; i < (1 << N) - 1; i++){
		for(j = 0; j < (1 << M) - 1; j++){
			printf("%d", a[i][j]);
		}
		printf("\n");
	}
	return 0;
}