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 is n ⋅ n!.

Let 1 ≤ i ≤ j ≤ n ⋅ n! be a pair of indices. We call the sequence (P_i, P_{i+1}, ..., P_{j-1}, P_j) a subarray of P. 

You are given n. Find the number of distinct subarrays of P. Since this number may be large, output it modulo 998244353 (a prime number). 

Input

The only line contains one integer n (1 ≤ n ≤ 10^6), as described in the problem statement.

Output

Output a single integer — the number of distinct subarrays, modulo 998244353.

Examples

Input


2


Output


8


Input


10


Output


19210869

Note

In the first example, the sequence P = [1, 2, 2, 1]. It has eight distinct subarrays: [1], [2], [1, 2], [2, 1], [2, 2], [1, 2, 2], [2, 2, 1] and [1, 2, 2, 1]. 
'''
// There are n! / (i+1)! * i permutations with decreasing seq length i (i+1 changes) for i <= n!
// There are n! / (i+1)! permutations for each 1 <= j <= i of position of moved one

// All nonempty things
+ n*n! * (n*n! + 1) / 2

// Subtract off everything inside one of them
- n! * n * (n+1) / 2

// Subtract off everything that doesn't include the min-incr
- sum_{perm i,j} n * (n-i)

// We're left with all words that include at least last of prv -> min-incr, which all have at least one match

// No matches
+ sum_{i=1..n} n!/(n-i)!

// First match is dist N, doesn't include anything
// aaa....aaa -> middle part can't all be decreasing
+ sum_{i=1..n} n! - n!/(n-i)!

// First match is dist N, includes subst
// aaabxyc   aaac
// Everything decreasing xycb
+ sum_{i=1..n-1} n!/(i+1)! * (n-i-1)

// First match is dist N, includes min-incr -> is unique, already counted

// First match is < N -> either came from min-incr or from subst
// You can distinguish because match..match includes everything smaller or everything smaller minus one of them
// Ending at min-incr gives unique thing

// How many distinct things end at subst?
// Ending at subst:   (b ijk ) lmn c xyz  ... c such that zyxbcnmlkji are sorted leftovers
                   or   b ijk lmn c xyz .... c
// Might as well not include xyz
+ sum_{i=1..n-1} n!/(i+1)! * (i+1)

// Simplified claim:
if it includes min-incr, then it's always unique;
if it includes subst, only count if it came from the end (j = 1) (in which case it always matches)
otherwise, it's just some delta-N match or it's all distinct

n*n! * (n*n! + 1) / 2
- n! * n * (n+1) / 2
- sum_{i=1..n-1} n! / (i+1)! * i * n * (n-i)
+ sum_{i=1..n} n! / (n-i)!
+ sum_{i=1..n} n! - n! / (n-i)!
+ sum_{i=1..n-1} n! / (i+1)! * n

= n*n! * (n*n! + 1) / 2
- n! * n * (n+1) / 2
- sum_{i=1..n-1} n! / (i+1)! * n * (i * (n-i) - 1)
+ n * n!

= n*n! * (n*n! - n + 2) / 2
- sum_{i=1..n-1} n! / (i+1)! * n * (i * (n-i) - 1)
'''

def fact(n):
    res = 1
    for i in range(1,n+1):
        res *= i
    return res

def f(n):
    l = fact(n) * n
    ans = l * (l-n+2) // 2
    for i in range(1,n):
        ans -= fact(n) // fact(i+1) * n * (i * (n-i) - 1)
    return ans

#print(f(int(input())) % 998244353)

def g(n):
    M = 998244353
    p = n
    a = 0
    for i in range(n,1,-1):
        a=(a+p*(i-1)*(n-i+1)-p)%M
        p=p*i%M
    a=(p*(p-n+2)-a-a)%M
    if a&1:a+=M
    return a//2

print(g(int(input())))
