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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 diff(self, order=1):
 """Differentiate a B-spline `order` number of times. Parameters: order: int, >= 0 Returns: The returned function internally uses __call__, which is 'memoized' for speed. """

        order = int(order)
        if order < 0:
            raise ValueError("order must be >= 0, got %d" % (order))

        if order == 0:
            return self.__call__

        if order > self.p:   # identically zero, but force the same output format as in the general case
            dummy = self.__call__(0.)  # get number of basis functions and output dtype
            nbasis = dummy.shape[0]
            return lambda x: np.zeros( (nbasis,), dtype=dummy.dtype )  # accept but ignore input x

        # At each differentiation, each term maps into two new terms.
        # The number of terms in the result will be 2**order.
        #
        # This will cause an exponential explosion in the number of terms for high derivative orders,
        # but for the first few orders (practical usage; >3 is rarely needed) the approach works.
        #
        terms = [ (1.,self) ]
        for k in range(order):
            tmp = []
            for Ci,Bi in terms:
                tmp.extend( (Ci*cn, Bn) for cn,Bn in Bi.__diff_internal() )  # NOTE: also propagate Ci
            terms = tmp

        # perform final summation at call time
        return lambda x: sum( ci*Bi(x) for ci,Bi in terms )