Output #1
Aoki
Initially, from the first pile at most $floor(5/2)=2$ stones can be removed at a time, and from the second pile at most $floor(3/3)=1$ stone can be removed at a time.
* If Takahashi first takes two stones from the first pile, from the first pile at most $floor(3/2)=1$ stone can now be removed at a time, and from the second pile at most $floor(3/3)=1$ stone can be removed at a time.
* Then, if Aoki takes one stone from the second pile, from the first pile at most $floor(3/2)=1$ stone can be removed at a time, and from the second pile no more stones can be removed (since $floor(2/3)=0$).
* Then, if Takahashi takes one stone from the first pile, from the first pile at most $floor(2/2)=1$ stone can now be removed at a time, and from the second pile no more stones can be removed.
* Then, if Aoki takes one stone from the first pile, from the first pile at most $floor(1/2)=0$ stones can now be removed at a time, and from the second pile no more stones can be removed.
No more operation can be performed, thus Aoki wins. If Takahashi plays differently, Aoki can also win by play accordingly.