he source iterable progresses, so beware of
wrapping very large or infinite iterables. Supply *maxlen* to limit the
size of the cache (this of course limits how far back you can seek).

    >>> from itertools import count
    >>> it = seekable((str(n) for n in count()), maxlen=2)
    >>> next(it), next(it), next(it), next(it)
    ('0', '1', '2', '3')
    >>> list(it.elements())
    ['2', '3']
    >>> it.seek(0)
    >>> next(it), next(it), next(it), next(it)
    ('2', '3', '4', '5')
    >>> next(it)
    '6'

Nc