What are the intermediate subfactors of the tensor product of two maximal subfactors?

This answer came after a discussion with Feng Xu. The following more general result is true:

Theorem: Let $(N_i \subset M_i)$, $i=1,2$, be irreducible finite index subfactors. Then $$ \mathcal{L}(N_1 \subset M_1) \times \mathcal{L}(N_2 \subset M_2) \subsetneq \mathcal{L}(N_1 \otimes N_2 \subset M_1 \otimes M_2)$$ if and only if they are intermediate subfactors $N_i \subseteq P_i \subset Q_i \subseteq M_i$, $i=1,2$, such that $(P_i \subset Q_i)$ is depth $2$ and isomorphic to $(R^{\mathbb{A}_i} \subset R)$, with $\mathbb{A}_2 \simeq \mathbb{A}_1^{cop}$ which is the Kac algebra $\mathbb{A}_1$ with the opposite coproduct.

Proof: This theorem was proved in the $2$-supertransitive case by Y. Watatani. The general case was conjectured by the OP, and proved by a discussion with Feng Xu as follows:
Let the intermediate subfactors $$N_1 \otimes N_2 \subseteq P_1 \otimes P_2 \subset R \subset Q_1 \otimes Q_2 \subseteq M_1 \otimes M_2$$ with $R$ not of tensor product form, $P_1 \otimes P_2$ and $Q_1 \otimes Q_2$ the closest (below and above resp.) to $R$ among them of tensor product form. Now using Proposition 3.5 (2) of [xu], $(P_i \subseteq Q_i)$, $i=1,2$, are depth $2$, there corresponding Kac algebras, $\mathbb{A}_i$, $i=1,2$, are very simple and $\mathbb{A}_2 \simeq \mathbb{A}_1^{cop}$ (see Definition 3.6 and Proposition 3.10 of [xu]). The converse is given by Theorem 3.14 (2) of [xu].