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This module implements certain functionality from PKCS#1 version 1.5. For a
very clear example, read http://www.di-mgt.com.au/rsa_alg.html#pkcs1schemes

At least 8 bytes of random padding is used when encrypting a message. This makes
these methods much more secure than the ones in the ``rsa`` module.

WARNING: this module leaks information when decryption fails. The exceptions
that are raised contain the Python traceback information, which can be used to
deduce where in the process the failure occurred. DO NOT PASS SUCH INFORMATION
to your users.
�N)�compare_digest�)�common�	transform�core�keys0 0*�H��
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    :return: 00 02 RANDOM_DATA 00 MESSAGE

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    >>> len(block)
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    >>> block[0:2]
    b'\x00\x02'
    >>> block[-6:]
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    The padding is always a repetition of FF bytes.

    :return: 00 01 PADDING 00 MESSAGE

    >>> block = _pad_for_signing(b'hello', 16)
    >>> len(block)
    16
    >>> block[0:2]
    b'\x00\x01'
    >>> block[-6:]
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    >>> block[2:-6]
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    :param pub_key: the :py:class:`rsa.PublicKey` to encrypt with.
    :raise OverflowError: when the message is too large to fit in the padded
        block.

    >>> from rsa import key, common
    >>> (pub_key, priv_key) = key.newkeys(256)
    >>> message = b'hello'
    >>> crypto = encrypt(message, pub_key)

    The crypto text should be just as long as the public key 'n' component:

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    :param crypto: the crypto text as returned by :py:func:`rsa.encrypt`
    :param priv_key: the :py:class:`rsa.PrivateKey` to decrypt with.
    :raise DecryptionError: when the decryption fails. No details are given as
        to why the code thinks the decryption fails, as this would leak
        information about the private key.


    >>> import rsa
    >>> (pub_key, priv_key) = rsa.newkeys(256)

    It works with strings:

    >>> crypto = encrypt(b'hello', pub_key)
    >>> decrypt(crypto, priv_key)
    b'hello'

    And with binary data:

    >>> crypto = encrypt(b'\x00\x00\x00\x00\x01', pub_key)
    >>> decrypt(crypto, priv_key)
    b'\x00\x00\x00\x00\x01'

    Altering the encrypted information will *likely* cause a
    :py:class:`rsa.pkcs1.DecryptionError`. If you want to be *sure*, use
    :py:func:`rsa.sign`.


    .. warning::

        Never display the stack trace of a
        :py:class:`rsa.pkcs1.DecryptionError` exception. It shows where in the
        code the exception occurred, and thus leaks information about the key.
        It's only a tiny bit of information, but every bit makes cracking the
        keys easier.

    >>> crypto = encrypt(b'hello', pub_key)
    >>> crypto = crypto[0:5] + b'X' + crypto[6:] # change a byte
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    ...
    rsa.pkcs1.DecryptionError: Decryption failed

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