How to understand canonical basis and dual canonical basis are dual to each other in the case of $sl_3$?

Question

Let $\mathfrak{g}=\mathfrak{n} \oplus \mathfrak{h} \oplus \mathfrak{n}^-$ be a triangular decomposition of a simple Lie algebra. Let $v$ be an indeterminant and let $U_v(\mathfrak{n})$ be the positive part of the quantum group $U_v(\mathfrak{g})$. In the case of $\mathfrak{g} = sl_3$, the canonical basis of $U_v(\mathfrak{n})$ is \begin{align*} B = \{E_1^{(a)} E_2^{(b)} E_1^{(c)}, E_2^{(a)} E_1^{(b)} E_2^{(c)}: b \ge a+c,a,c \ge 0\}, \end{align*} where the elements $E_1^{(a)} E_2^{(a+c)} E_1^{(c)} = E_2^{(c)} E_1^{(a+c)} E_2^{(a)}$ are considered only once. On the other hand, the dual canonical basis of $\mathbb{C}[N]$, $N$ is the unipotent subgroup of $SL_3$ consisting of all unipotent upper triangular matrices. The dual canonical basis of $\mathbb{C}[N]$ is \begin{align*} B^* = \{x_{12}^a x_{13}^b (x_{12}x_{23}-x_{13})^c, x_{23}^a x_{13}^b (x_{12}x_{23}-x_{13})^c: a,b,c \ge 0 \}. \end{align*} Let $A_v(\mathfrak{n})$ be the graded dual of $U_v(\mathfrak{n})$. How to write the dual canonical basis $\mathbf{B}^*$ of $A_v(\mathfrak{n})$ explicitly? I am trying to understand how this basis $\mathbf{B}^*$ specialize to $B^*$ when $v \to 1$. Thank you very much.