Is S-T CONNECTEDNESS #P-complete on instances when all s-t paths are of the same length? S-T CONNECTEDNESS

Input: a (undirected) graph $G=(V,E)$; $s,t \in V.$

Output: number of spanning subgraphs of $G$ in which there is a path from $s$ to $t$.



S-T CONNECTEDNESS problem is known to be #P-complete ([Valiant, L. G. (1979). The complexity of enumeration and reliability problems. SIAM Journal on Computing, 8(3), 410-421.]).

I'm interested in those instances of S-T CONNECTEDNESS in which all simple paths between $s$ and $t$ are of the same length. Is anything known about the complexity of the problem on these instances? In particular, does the problem remain #P-complete?

Remark: In the Valiant's reduction, there are many simple paths between $s$ and $t$ of different lengths.
After polynomial time preprocessing, you may assume that every edge in $E$ lies on some simple $s$-$t$-path.


If all simple paths between $s$ and $t$ are of the same length, then $G$ is series-parallel, with $s$ as left terminal and $t$ as right terminal vertex. This can be seen easily, since the forbidden subgraphs for series-parallel graphs (= subdivisions of diamonds) create two paths of different lengths between $s$ and $t$.
In a series-parallel graph, the number of spanning subgraphs containing a simple $s$-$t$-path can be determined in polynomial time by dynamic programming. 


In the dynamic program, we build up the graph by series and parallel compositions. For every resulting subgraph $G'$, we compute the number $p^+(G')$ of spanning subgraphs that do contain a simple path from its left terminal to its right terminal and the number $p^-(G')$ of spanning subgraphs that do not contain such a path. 


In a series composition on $G_1$ and $G_2$, we get $p^+(G)=p^+(G_1)\cdot p^+(G_2)$.
In a parallel composition on $G_1$ and $G_2$, we get $p^-(G)=p^-(G_1)\cdot p^-(G_2)$.
In both cases, the missing value $p^+(G)$ or $p^-(G)$ for $G=(V,E)$ can be computed from $p^+(G)+p^-(G)=2^{|E|}$.