Laertes and Roxane are tired of cutting cakes. They want to eat some now, and they do so by playing a game as follows:

On his turn, Laertes must choose any square on the grid and eat it, along with the square immediately below it. On her turn, Roxane must choose any square on the grid and eat it, along with the square to its immediate right. The first person who cannot eat two pieces as described above on any turn loses. Let us describe the squares using the conventional matrix coordinate scheme, with the top-left square being $(0,0)$ and the positive $x$ and $y$ axes extend to the right and down respectively. In cases where taking two different squares results in the same cake upto symmetry, then the square that is closest to the origin is preferred. If the square is still ambiguous, then the one that is closer to the $x$ axis is preferred.

Consider the $2 \times 1$ cake. It is obvious that no matter who goes first, Laertes can always win by eating the square at $(0,0)$ (and so consequently $(1,0)$).

The $2 \times 2$ cake is different though. Convince yourself that the first person to start will win, no matter who it is.

What are the optimal squares chosen by both players on a $4 \times 2$ cake? Assume that Laertes goes first. If you think the sequence is $(a,b)$, $(c,d)$, $(e,f)$, enter $abcdef$.