#P1091D. New Year and the Permutation Concatenation
New Year and the Permutation Concatenation
Description
Let $n$ be an integer. Consider all permutations on integers $1$ to $n$ in lexicographic order, and concatenate them into one big sequence $p$. For example, if $n = 3$, then $p = [1, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 3, 1, 2, 3, 2, 1]$. The length of this sequence will be $n \cdot n!$.
Let $1 \leq i \leq j \leq n \cdot n!$ be a pair of indices. We call the sequence $(p_i, p_{i+1}, \dots, p_{j-1}, p_j)$ a subarray of $p$. Its length is defined as the number of its elements, i.e., $j - i + 1$. Its sum is the sum of all its elements, i.e., $\sum_{k=i}^j p_k$.
You are given $n$. Find the number of subarrays of $p$ of length $n$ having sum $\frac{n(n+1)}{2}$. Since this number may be large, output it modulo $998244353$ (a prime number).
The only line contains one integer $n$ ($1 \leq n \leq 10^6$), as described in the problem statement.
Output a single integer — the number of subarrays of length $n$ having sum $\frac{n(n+1)}{2}$, modulo $998244353$.
Input
The only line contains one integer $n$ ($1 \leq n \leq 10^6$), as described in the problem statement.
Output
Output a single integer — the number of subarrays of length $n$ having sum $\frac{n(n+1)}{2}$, modulo $998244353$.
Samples
3
9
4
56
10
30052700
Note
In the first sample, there are $16$ subarrays of length $3$. In order of appearance, they are:
$[1, 2, 3]$, $[2, 3, 1]$, $[3, 1, 3]$, $[1, 3, 2]$, $[3, 2, 2]$, $[2, 2, 1]$, $[2, 1, 3]$, $[1, 3, 2]$, $[3, 2, 3]$, $[2, 3, 1]$, $[3, 1, 3]$, $[1, 3, 1]$, $[3, 1, 2]$, $[1, 2, 3]$, $[2, 3, 2]$, $[3, 2, 1]$.
Their sums are $6$, $6$, $7$, $6$, $7$, $5$, $6$, $6$, $8$, $6$, $7$, $5$, $6$, $6$, $7$, $6$. As $\frac{n(n+1)}{2} = 6$, the answer is $9$.