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
// 解き直し.
// https://img.atcoder.jp/hitachi2020/editorial.pdf
// C++(GCC 9.2.1)
#include <bits/stdc++.h>
using namespace std;
#define repex(i, a, b, c) for(int i = a; i < b; i += c)
#define repx(i, a, b) repex(i, a, b, 1)
#define rep(i, n) repx(i, 0, n)
#define repr(i, a, b) for(int i = a; i >= b; i--)
const int MAX = 1 << 10;
int gCum[MAX][MAX], grid[MAX][MAX];

int main(){
    
    // 1. 入力情報.
    int N, M;
    scanf("%d %d", &N, &M);
    
    // 2. 大きい方を保存.
    int eMax = max(N, M);
    
    // 以下, 解説通り.
    // https://www.youtube.com/watch?v=z9zEs79hHPE
    // 3. 累積和作成.
    rep(i, 1 << eMax) rep(j, 1 << eMax) gCum[i][j] = __builtin_popcount(i & j) % 2;
    
    // 4. 累積和からグリッド復元.
    rep(i, (1 << eMax) - 1){
        rep(j, (1 << eMax) - 1){
            grid[i][j] = gCum[i + 1][j + 1] + gCum[i + 1][j] + gCum[i][j + 1] + gCum[i][j];
            grid[i][j] %= 2;
        }
    }
    
    // 5. 出力.
    rep(i, (1 << N) - 1){
        rep(j, (1 << M) - 1) printf("%d", grid[i][j]);
        puts("");
    }
    return 0;
    
}