D. New Year and the Permutation Concatenationtime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputLet 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). InputThe only line contains one integer n (1 \leq n \leq 10^6), as described in the problem statement.OutputOutput a single integer — the number of subarrays of length n having sum \frac{n(n+1)}{2}, modulo 998244353.ExamplesInput
3
Output
9
Input
4
Output
56
Input
10
Output
30052700
NoteIn 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.
3

9 
['combinatorics', 'dp', 'math', '*1700']