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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 fold_xor(bloomfilter, # type: bitarray folds # type: int ):
 """ Performs XOR folding on a Bloom filter. If the length of the original Bloom filter is n and we perform r folds, then the length of the resulting filter is n / 2 ** r. :param bloomfilter: Bloom filter to fold :param folds: number of folds :return: folded bloom filter """


    if len(bloomfilter) % 2 ** folds != 0:
        msg = ('The length of the bloom filter is {length}. It is not '
               'divisible by 2 ** {folds}, so it cannot be folded {folds} '
               'times.'
               .format(length=len(bloomfilter), folds=folds))
        raise ValueError(msg)

    for _ in range(folds):
        bf1 = bloomfilter[:len(bloomfilter) // 2]
        bf2 = bloomfilter[len(bloomfilter) // 2:]

        bloomfilter = bf1 ^ bf2

    return bloomfilter