2 regular graphs and permutations I have found this question on MSE before but I didn't find the answer satisfactory and it is so old I doubt anyone is still following it. 
Let $f_{n}$ be the number of permutations on $[n]$ with no fixed points or two cycles. Let $g_{n}$ be the number of simple, labeled two regular graphs on $n$ vertices. Let $F(x)$ be the EGF for $f_{n}$ and $G(x)$ be the EGF for $g_{n}$. I've shown that $F(x)=G(x)^{2}$. So by composition of exponential genereating functions 
$$f_{n}=\sum_{k\geq 0}^{n}\binom{n}{k}g_{k}g_{n-k}$$. 
I am tasked to prove $F(x)=G(x)^{2}$ via a bijective proof. I know that given a graph on $k$ vertices we can decompose it into cycles. Then for each cycle we can orient the cycle two ways to make a permutation on $n$. But I think this count is too large. 
Any hint would help thanks.   

HINT: Each permutation of $[n]$ with no fixed points and no $2$-cycles corresponds in an obvious way to a unique labelled two-regular directed graph in which each cyclic component is a directed cycle. Given such a graph $G$, we can split the components into two sets, $\mathscr{C}_I(G)$ and $\mathscr{C}_D(G)$, as follows:

Suppose that $C$ is a component whose lowest-numbered vertex is $v$; there are vertices $u$ and $w$ in $C$ such that $u\to v$ and $v\to w$ are directed edges of $C$. If $u<w$, put $C$ into $\mathscr{C}_I(G)$, and if $u>w$, put $C$ into $\mathscr{C}_D(G)$.

Let $G_I=\bigcup\mathscr{C}_I$ and $G_D=\bigcup\mathscr{C}_D$. Now show that for a fixed set of $k$ labels from $[n]$ there are $g_k$ possible graphs $G_I$ and $g_{n-k}$ possible graphs $G_D$.
