Optimized Mermin equals optimized Bell

Let $Q_M$ be the table for two parties with three settings each and binary outcomes, whose value for each equal-outcome pair is $1/2$ when the settings agree and $1/8$ when they differ, and whose value for each unequal-outcome pair is $0$ when the settings agree and $3/8$ when they differ. Let $Q_B$ be the table for two parties with two settings each and binary outcomes, whose equal-outcome/unequal-outcome values are $(1/2,0)$ at setting $(0,0)$, $(1/8,3/8)$ at setting $(0,1)$, and $(3/8,1/8)$ at settings $(1,0)$ and $(1,1)$, with each value assigned to each individual outcome pair of that kind. For either scenario, let $A$ be its setting space, $V$ the space of binary assignments $v_{ia}$ to every party and every one of its settings, $\Delta_A,\Delta_V$ their probability simplices, and $\mathcal U\subseteq\Delta_A$ the distributions factoring into single-party setting distributions. For $p\in\Delta_V$, define $L_p(a,x)=\sum_{v:\,\forall i,\ v_{i,a_i}=x_i}p(v)$. Define $K_Q(s,p)$ from $q(a,x)=s(a)Q(a,x)$ and $r(a,x)=s(a)L_p(a,x)$ as $+\infty$ if some $q(a,x)>0$ has $r(a,x)=0$, and otherwise as the nonnegative part of $\sum_{a,x}q(a,x)\log_2(q(a,x)/r(a,x))$, with zero-$q$ summands set to $0$. Define

as extended nonnegative real numbers. The declaration asserts $U(Q_M)=U(Q_B)$ and $C(Q_M)=C(Q_B)$.