552 「LibreOJ Round #8」MIN&MAX I

For an $n$-order permutation $p$, we set up an undirected simple graph $G(p)$ with $n$ vertices numbered from $1$ to $n$. We create an edge between each vertice $i$ and the nearest vertices in each side which correspond a greater (or less) $p$ value than $p_i$.
Formally,in this graph, $\forall u<v$, the edge $(u, v)$ exists iff at least one of the following four conditions hold:

• $p_u<p_v$, and no $u<i<v$ exists such that $p_u<p_i$;
• $p_u>p_v$, and no $u<ip_i$;
• $p_u<p_v$, and no $u<i<v$ exists such that $p_i<p_v$;
• $p_u>p_v$, and no $u<ip_v$.

Now we randomly choose a permutation $p$ from all $n$-order permutations. Your task is to calculate the expected number of the $3$-cycles in $G(p)$. You only need to output the answer modulo $998244353$.

The only line contains a positive integer $n$ which means the order of the permutation.

Output only one line,which contains an integer $\mathrm{ans}$ which means the expected number of the $3$-cycles in $G(p)$ modulo $998244353$.

It is easy to count that there are four $3$-cycles in total from the $3!$ permutations(each of ${1,3,2},{2,3,1},{2,1,3},{3,1,2}$ has one). So answer is $\frac{4}{3!}=\frac{2}{3}$,that is, $2\times 3^{-1} \pmod{998244353}=665496236$.

For all test cases, $1\le n<998244353$.

Detailed constraints are as follows (blank grids denote the same constraints as mentioned above):