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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 pick(self, connections):
 """Picks a connection with the earliest backoff time. As a result, the first connection is picked for as long as it has no backoff time. Otherwise, the connections are tried in a round robin fashion. Args: connections (:obj:list):
 List of :class:`~bigchaindb_driver.connection.Connection` instances. """

        if len(connections) == 1:
            return connections[0]

        def key(conn):
            return (datetime.min
                    if conn.backoff_time is None
                    else conn.backoff_time)

        return min(*connections, key=key)