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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def group_by_count(iterable: List[Any], count: int, default_value: Any) -> List[List[Any]]: """ Takes a list and groups it into sublists of size ``count``, using ``default_value`` to pad the list at the end if the list is not divisable by ``count``. For example: [[1, 2, 3], [4, 5, 6], [7, 0, 0]] This is a short method, but it's complicated and hard to remember as a one-liner, so we just make a function out of it. """

    return [list(l) for l in zip_longest(*[iter(iterable)] * count, fillvalue=default_value)]