Let $n \in \mathbb{N}$, $n \geq 3$ and $n = 2m$ for some $m \in \mathbb{N}$. How do I efficently get the result of the product of cycles $$\begin{pmatrix} 2 & n\end{pmatrix}\begin{pmatrix} 3 & n-1\end{pmatrix}\cdots\begin{pmatrix} m & m+2\end{pmatrix}\begin{pmatrix} 1 & 2 & \dots & n\end{pmatrix}\begin{pmatrix} 2 & n\end{pmatrix}\begin{pmatrix} 3 & n-1\end{pmatrix}\cdots\begin{pmatrix} m & m+2\end{pmatrix}$$
I could write it somehow in the typical form $$\begin{pmatrix} 1 & 2 & \dots & m & m+1 & m+2 & \dots & n-1 & n\\ 1 & 2 & \dots & m & m+1 & m+2 & \dots & n-1 & n\end{pmatrix}$$ but this gets really messy. Thanks.
Edit. An idea would be induction on $m$.
We have $(k \; n-k+2)\; (n-k+2 \; n-k+3)\; (n-k+3 \;k-1) = (k \; k-1) $ (a middle transposition comes from a $(1 \; 2 \; \ldots \; n)$ cycle ) and it could be applied to compute this product:
$k$ goes to $k-1$ hence we have a cycle:
$ (n\; n-1 \; n-2\; \ldots \; 3 \; 2 \; 1)$.