Let us proceed to the unrestricted T-systems. Choose $h\in {\mathbb{C}\backslash 2\pi \sqrt{-1} \mathbb{Q}}$ arbitrarily.
The unrestricted T-system for $U_{q}(X_{N}^{(\mathfrak{k})})$ is the following relations.
For $X_{N}^{(\mathfrak{k})}=A_{2r}^{(2)}$,
\begin{align}
T_{m}^{(a)}(u-1)T_{m}^{(a)}(u+1)&=T_{m-1}^{(a)}(u)T_{m+1}^{(a)}(u)+T_{m}^{(a-1)}(u)T_{m}^{(a+1)}(u), (1\leq a\leq r-1)\\
T_{m}^{(r)}(u-1)T_{m}^{(r)}(u+1)&=T_{m-1}^{(r)}(u)T_{m+1}^{(r)}(u)+T_{m}^{(r-1)}(u)T_{m}^{(r)}(u+\Omega),
\end{align}
where $\Omega=2\pi \sqrt{-1}/{\mathfrak{k}h}$.

My question is how to value $\Omega$ in T-system for twisted quantum affine algebras?

For example in type $A_{r}$, the untwisted case.
\begin{align} 
T_{m}^{(a)}(u-1)T_{m}^{(a)}(u+1)&=T_{m-1}^{(a)}(u)T_{m+1}^{(a)}(u)+T_{m}^{(a-1)}(u)T_{m}^{(a+1)}(u).
 \end{align}
$A_{3}$, if $a=m=2$, then 
\begin{align}
2_{-4}2_{-2}\ast2_{-2}2_{0}=2_{-2}*2_{-4}2_{-2}2_{0}+1_{-3}1_{-1}*3_{-3}3_{-1}
\end{align}
then we have
\begin{align}
T_{2}^{(2)}(-3)T_{2}^{(2)}(-1)&=T_{1}^{(2)}(-2)T_{3}^{(2)}(-2)+T_{2}^{(1)}(-2)T_{2}^{(3)}(-2),
 \end{align}
which satisfies the above equation.
I want to get the similar equation like $A_{3}$