MAC unforgeability and secret key guessing The standard notion security for MACs is usually expressed by means of experiments like the following one (given an attacker A):


$k  \leftarrow \mathsf{KG}(1^\lambda)$
Invoke $A^{\mathsf{TAG}_k(\cdot),\mathsf{VRFY}_k(\cdot,\cdot)}$
Output $1$ if $A$ queried $(m^*,\tau^*)$ to $\mathsf{VRFY}_k(\cdot,\cdot)$ s.t.  $\mathsf{VRFY}_k(m^*,\tau^*) = \mathsf{Accept}$ and $A$ didn't receive $\tau^*$ by querying  $m^*$ to $\mathsf{TAG}_k(\cdot)$


Now consider the following variation to point 3:
Output $1$ if $A$ gives $k'$ s.t. $k=k'$

How could I prove that a MAC unforgeable according to the first game is unforgeable according to the second game too? Is there a specific $\mathsf{MAC}(\mathsf{KG},\mathsf{TAG},\mathsf{VRFY})$ that is unforgeable according to the second game, but not according to the first one?
I was going to make this a comment; however you asked for hint, and these are hints.

Suppose you had an Oracle that solved the second game for you; how could you use that Oracle to solve the first game?

Does that imply that a MAC where the first game is unsolvable imply that the second game is also unsolvable?