Fierz identity for symplectic group

For the fundamental representation of $$SU(N)$$, there is a Fierz identity: $$\sum_iT^i_{ab}T^i_{cd}=\frac{1}{2}\left(\delta_{ad}\delta_{bc}-\frac{1}{N}\delta_{ab}\delta_{cd}\right)$$ where $$T^i$$ is the $$i$$th generator of $$SU(N)$$ normalized as $${\rm Tr}(T^iT^j)=\frac{1}{2}\delta^{ij}.$$

My question is: For the symplectic group $$Sp(N)$$, what is the Fierz identity? Namely, suppose $$T^i$$ is the $$i$$th generator of $$Sp(N)$$, can $$\sum_iT^i_{ab}T^i_{cd}$$ be written as something like the identity above?

The generators of $$\mathrm{Sp}(2n)$$ may be taken as matrices
T^a = \left\lbrace\begin{aligned} &A^a & 1 \leq &a \leq n^2 \\ &B^a & n^2+1 \leq &a \leq \tfrac12n(3n + 1) \\ &C^a & \tfrac32n(n + 1) \leq &a \leq n(2n + 1) \end{aligned}\right. where $$A^a,B^a$$ and $$C^a$$ are defined as $$2\times 2$$ block matrices with $$n\times n$$ blocks $$A^a=\left(\begin{matrix}m^a & 0 \\0& -m^{a\,\mathsf{T}}\end{matrix}\right)\,,\qquad B^a=\left(\begin{matrix}0 & s^{a-n^2} \\0& 0\end{matrix}\right)\,,\qquad C^a=\left(\begin{matrix}0 & 0 \\s^{a-\frac12n(3n+1)}& 0\end{matrix}\right)\,.$$ with $$m^a$$ being a matrix with all zeros and a $$1$$ in the $$a$$th entry and $$s^a$$ being the $$a$$th matrix in the basis of symmetric matrices. If we let $$a$$ be a multi index $$a \to (\mu,\nu)$$ we can write those generators explicitly in terms of Kronecker deltas \begin{aligned} (m^{\mu\nu})_{ij} &= \delta^\mu_{i} \delta^\nu_j\,,\\ (s^{\mu\nu})_{ij} &= \tfrac{1}{2\sqrt{2}}\left(\delta^\mu_{i} \delta^\nu_j+\delta^\mu_{j} \delta^\nu_i\right)\,. \end{aligned} I'm not entirely sure about the normalization.$${}^1$$ Now you can write your sum as (sorry if I named the indices differently) $$\sum_{a=1}^{n(2n+1)} (T^a)_{IJ}\,(T^a)_{KL} = \sum_{\mu,\nu = 1}^n (T^{\mu\nu})_{IJ}\,(T^{\mu\nu})_{KL}\,,$$ and now you'll have many cases according to whether $$I,J,K,L$$ is smaller or larger than $$n$$. If it's smaller then $$I\to i$$ and if it's bigger $$I \to i - n$$, and the appropriate blocks need to be considered. In the end you'll end up with contractions involving only $$\delta^\mu_i$$ which can be done easily for general $$N = 2n$$.
$$\qquad{}^1$$ Anyway you can trace everything in the end and see if it's correct.