import sys

sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI1(): return list(map(int1, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def SI(): return sys.stdin.readline().rstrip()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = 10**16
md = 998244353
# md = 10**9+7

class BitSum:
    def __init__(self, n):
        self.n = n+1
        self.table = [0]*self.n

    def add(self, i, x):
        i += 1
        while i < self.n:
            self.table[i] += x
            i += i & -i

    # [0,i]の和
    def sum(self, i):
        i += 1
        res = 0
        while i > 0:
            res += self.table[i]
            i -= i & -i
        return res

    # [l,r)の和
    def sumlr(self, l, r):
        if l >= r: return 0
        if l == 0: return self.sum(r-1)
        return self.sum(r-1)-self.sum(l-1)

    # 数列を度数分布とみたときに、x番目がどのインデックスにあるかを返す
    # xが大きすぎるときは配列の長さnを返す
    def rank(self, x):
        idx = 0
        for lv in range((self.n-1).bit_length()-1, -1, -1):
            mid = idx+(1 << lv)
            if mid >= self.n: continue
            if self.table[mid] < x:
                x -= self.table[mid]
                idx += 1 << lv
        return idx

def nHr(hn, hr):
    return nCr(hn+hr-1, hr-1)

def nPr(com_n, com_r):
    if com_r < 0: return 0
    if com_n < com_r: return 0
    return fac[com_n]*ifac[com_n-com_r]%md

def nCr(com_n, com_r):
    if com_r < 0: return 0
    if com_n < com_r: return 0
    return fac[com_n]*ifac[com_r]%md*ifac[com_n-com_r]%md

# 準備
n_max = 200005
fac = [1]
for i in range(1, n_max+1): fac.append(fac[-1]*i%md)
ifac = [1]*(n_max+1)
ifac[n_max] = pow(fac[n_max], md-2, md)
for i in range(n_max-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%md

n = II()
aa = LI()
enc = {a: i for i, a in enumerate(sorted(set(aa)))}
aa = [enc[a] for a in aa]
bit = BitSum(n)
# Aの転倒数
inv = 0
for i, a in enumerate(aa):
    inv += i-bit.sum(a)
    bit.add(a, 1)
# 位置は関係なく大小のペア
pair = 0
for a in aa:
    pair += n-bit.sum(a)
# 同グループ内に、ある転倒数ができる期待値
ev1 = 0
for d in range(2, n+1):
    ev1 += nCr(n-2, d-2)*pow(nCr(n, d), md-2, md)%md
    ev1 %= md
# 転倒数の個数をかける
ev1 = ev1*inv%md
# 別グループ間で、ある転倒数ができる期待値
ev2 = 0
s = 0
for d in range(1, n+1):
    ev2 += s*d
    ev2 %= md
    s += d
ev2 = ev2*pow(n, 2*(md-2), md)%md
# ペアの数をかける
ev2 = ev2*pair%md
# すべての選び方を求める
tot = 1
for d in range(1, n+1):
    tot = tot*nCr(n, d)%md
# 期待値×場合の数で答え
ans = (ev1+ev2)*tot%md
print(ans)