## Task

Implement a function which finds the numbers less than `2`, and the indices of numbers greater than `1` in the given sequence, and returns them as a pair of sequences. 

Return a nested array or a tuple depending on the language:

* The first sequence being only the `1`s and `0`s from the original sequence. 
* The second sequence being the indexes of the elements greater than `1` in the original sequence.  

## Examples

```python
[ 0, 1, 2, 1, 5, 6, 2, 1, 1, 0 ] => ( [ 0, 1, 1, 1, 1, 0 ], [ 2, 4, 5, 6 ] )
```

Please upvote and enjoy!

Difficulty: EASY
Starter code: def binary_cleaner(seq):
	

Solutions:
def binary_cleaner(seq):
	res = ([], [])
	for (i, x) in enumerate(seq):
		if x < 2:
			res[0].append(x)
		else:
			res[1].append(i)
	return res


def binary_cleaner(lst):
	return ([n for n in lst if n < 2], [i for (i, n) in enumerate(lst) if n > 1])


def binary_cleaner(seq):
	return (list(filter(lambda x: x <= 1, seq)), [i for i in range(len(seq)) if seq[i] > 1])


def binary_cleaner(seq):
	return ([x for x in seq if x in [0, 1]], [i for (i, x) in enumerate(seq) if x > 1])


def binary_cleaner(seq):
	return ([value for value in seq if value == 0 or value == 1], [index for (index, value) in enumerate(seq) if value not in (0, 1)])


def binary_cleaner(seq):
	return ([n for n in seq if n < 2], [i for (i, n) in enumerate(seq) if n > 1])


def binary_cleaner(seq):
	(list_a, list_b) = ([], [])
	for (k, v) in enumerate(seq):
		if v > 1:
			list_b.append(k)
		else:
			list_a.append(v)
	return (list_a, list_b)


def binary_cleaner(seq):
	b = []
	x = []
	for i in range(len(seq)):
		if seq[i] < 2:
			b.append(seq[i])
		elif seq[i] > 1:
			x.append(i)
	return (b, x)


def binary_cleaner(seq):
	less_than_1 = []
	index_list = []
	for (i, x) in enumerate(seq):
		if x < 2:
			less_than_1.append(x)
		if x > 1:
			index_list.append(i)
	return (less_than_1, index_list)


def binary_cleaner(seq):
	return ([i for i in seq if i <= 1], [i for (i, el) in enumerate(seq) if el > 1])