# Is CMAC vulnerable to length extension attacks?

Given $m_1, m_2$ such that $MAC_k(m_1) = MAC_k(m_2)$ is it possible to construct more collisions with pairs of the form $m_1|x, m_2|x$?

Here is the CMAC picture from Wikipedia where it is only part of the one-key MAC article:

I don't see how it could be because it treats the last block of the message differently but this was mentioned in the comments of another question so I wanted to clarify.

Here is the quote from poncho:

CMAC and XCBC are similar to the UMAC case (even though a universal hash is not involved); if you find two messages that have the same CMAC value, it is likely that a simple length-extension to the two messages will allow the attacker to create other pairs of messages with the same MAC.

Bonus question: Are there other ways to exploit a collision?