What is the number of local deterministic behaviours in a Bell scenario?

Consider a Bell scenario: Alice and Bob perform measurements on a shared system, and each of them has a "measurement box" with a bunch of inputs (corresponding to different measurement settings) and corresponding outputs.

Say they can each choose from $$m$$ possible inputs, labelled as $$x,y\in\{1,...,m\}$$, and can get one of $$\Delta$$ possible outputs $$a,b\in\{1,...,\Delta\}$$ (taking the notation from (1303.2849)).

In this context, a behaviour is a set of joint probabilities $$\boldsymbol p\equiv \{p(ab|xy)\}_{abxy}\in\mathbb R^{\Delta^2 m^2}$$. A local deterministic behaviour is a behaviour such that $$p(ab|xy)=\delta_{a,a_x}\delta_{b,b_x}$$ for some pair $$(a_x,b_y)$$ associated with each $$(x,y)$$.

I want to count the number of such behaviours.

To do this, I would have said that, for each $$(x,y)$$, there are $$\Delta^2$$ possible outcome assignments, as $$a_x$$ and $$b_x$$ can be chosen independently. At the same time, there are $$m^2$$ possible pairs $$(x,y)$$, and thus the total number should be $$(\Delta^2)^{m^2}=\Delta^{2 m^2}$$. However, in (1303.2849) (between (17) and (18)) they state that this number is $$\Delta^{2m}$$. Why is my calculation off?

The problem with the calculation is that it was not taking into account the locality constraint. Namely, the fact that local deterministic behaviours also have to satisfy $$p(ab|xy)=p(a|x)p(b|y).$$ Taking this into account, we notice that there are $$\Delta^m$$ possible assignments for $$p(a|x)$$, and the same number of $$p(b|y)$$, making for a total number of $$(\Delta^m)^2=\Delta^{2m}$$ local deterministic assignments, as reported in the reference.