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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 ciphertext(self, be_secure=True):
 """Return the ciphertext of the EncryptedNumber. Choosing a random number is slow. Therefore, methods like :meth:`__add__` and :meth:`__mul__` take a shortcut and do not follow Paillier encryption fully - every encrypted sum or product should be multiplied by r ** :attr:`~PaillierPublicKey.n` for random r < n (i.e., the result is obfuscated). Not obfuscating provides a big speed up in, e.g., an encrypted dot product: each of the product terms need not be obfuscated, since only the final sum is shared with others - only this final sum needs to be obfuscated. Not obfuscating is OK for internal use, where you are happy for your own computer to know the scalars you've been adding and multiplying to the original ciphertext. But this is *not* OK if you're going to be sharing the new ciphertext with anyone else. So, by default, this method returns an obfuscated ciphertext - obfuscating it if necessary. If instead you set `be_secure=False` then the ciphertext will be returned, regardless of whether it has already been obfuscated. We thought that this approach, while a little awkward, yields a safe default while preserving the option for high performance. Args: be_secure (bool):
 If any untrusted parties will see the returned ciphertext, then this should be True. Returns: an int, the ciphertext. If `be_secure=False` then it might be possible for attackers to deduce numbers involved in calculating this ciphertext. """

        if be_secure and not self.__is_obfuscated:
            self.obfuscate()

        return self.__ciphertext