An onion array is an array that satisfies the following condition for all values of `j` and `k`:

If all of the following are `true`:

* `j >= 0`
* `k >= 0`
* `j + k = array.length - 1`
* `j != k`
  
then:

* `a[j] + a[k] <= 10`

### Examples:

```
[1, 2, 19, 4, 5]   =>  true  (as 1+5 <= 10 and 2+4 <= 10)
[1, 2, 19, 4, 10]  =>  false (as 1+10 > 10)
```

Write a function named `isOnionArray`/`IsOnionArray`/`is_onion_array()` that returns `true` if its argument is an onion array and returns `false` if it is not.

~~~if:php
Your solution should at least be moderately efficient.  Make sure you don't do any unnecessary looping ;)
~~~

Difficulty: EASY
Starter code: def is_onion_array(a):
	

Solutions:
def is_onion_array(a):
	return all((a[i] + a[-i - 1] <= 10 for i in range(len(a) // 2)))


from typing import List

def is_onion_array(a: List[int]) -> bool:
	return all((aj + ak <= 10 for (aj, ak) in zip(a, a[:(len(a) - 1) // 2:-1])))


def is_onion_array(a):
	for i in range(0, len(a) // 2):
		if a[i] + a[-i - 1] > 10:
			return False
	return True


def is_onion_array(a):
	for j in range(len(a) // 2):
		if a[j] + a[-j - 1] > 10:
			return False
	return True


def is_onion_array(a):
	if a == []:
		return True
	a = [sum(x) for x in zip(a, a[::-1])]
	a.pop(len(a) // 2)
	for element in a:
		if element > 10:
			return False
	return True


def is_onion_array(a):
	i = 0
	j = -1
	while i != len(a) // 2:
		if a[i] + a[j] > 10:
			return False
		i += 1
		j -= 1
	return True


def is_onion_array(a):
	print(a)
	for num in range(len(a)):
		if a[num] != a[-(num + 1)]:
			if not a[num] + a[-(num + 1)] <= 10:
				return False
	return True


from itertools import combinations as cb

def is_onion_array(a):
	return not any([sum(c) == len(a) - 1 and a[c[0]] + a[c[1]] > 10 for c in cb([*range(0, len(a))], 2)])