p02751 Social Infrastructure Information Systems Division  Hitachi Programming Contest 2020 - Odd Sum Rectangles
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
```python
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from itertools import permutations
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    def hadamard(n):
        if mem[n] != None:
            return mem[n]
        h = hadamard(n-1)
        res = [[j for j in i] for i in h]
        res_ = [[j for j in i] for i in h]
        for i in range(len(res)):
            res[i] += [j for j in h[i]]
            res_[i] += [j^1 for j in h[i]]
        res += res_
        mem[n] = res
        return res

    n,m = LI()
    f = 0
    if m < n:
        n,m = m,n
        f = 1
    h,w = 1<<n, 1<<m
    mem = defaultdict(lambda : None)
    mem[0] = [[0]]
    s = hadamard(m)[:h]
    if f:
        s = [[s[j][i] for j in range(h)] for i in range(w)]
        h,w = w,h
    for x in range(w):
        for y in range(h-1)[::-1]:
            s[y+1][x] ^= s[y][x]
    for y in range(h):
        for x in range(w-1)[::-1]:
            s[y][x+1] ^= s[y][x]
    ans = [i[1:] for i in s[1:]]
    for i in ans:
        print(*i,sep="")
    return

#Solve
if __name__ == "__main__":
    solve()

```