Your program fails again. This time it gets "Wrong answer on test 233".

This is the harder version of the problem. In this version, $1 \le n \le 2\cdot10^5$. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems.

The problem is to finish $n$ one-choice-questions. Each of the questions contains $k$ options, and only one of them is correct. The answer to the $i$-th question is $h_{i}$, and if your answer of the question $i$ is $h_{i}$, you earn $1$ point, otherwise, you earn $0$ points for this question. The values $h_1, h_2, \dots, h_n$ are known to you in this problem.

However, you have a mistake in your program. It moves the answer clockwise! Consider all the $n$ answers are written in a circle. Due to the mistake in your program, they are shifted by one cyclically.

Formally, the mistake moves the answer for the question $i$ to the question $i \bmod n + 1$. So it moves the answer for the question $1$ to question $2$, the answer for the question $2$ to the question $3$, ..., the answer for the question $n$ to the question $1$.

We call all the $n$ answers together an answer suit. There are $k^n$ possible answer suits in total.

You're wondering, how many answer suits satisfy the following condition: after moving clockwise by $1$, the total number of points of the new answer suit is strictly larger than the number of points of the old one. You need to find the answer modulo $998\,244\,353$.

For example, if $n = 5$, and your answer suit is $a=[1,2,3,4,5]$, it will submitted as $a'=[5,1,2,3,4]$ because of a mistake. If the correct answer suit is $h=[5,2,2,3,4]$, the answer suit $a$ earns $1$ point and the answer suite $a'$ earns $4$ points. Since $4 > 1$, the answer suit $a=[1,2,3,4,5]$ should be counted.


-----Input-----

The first line contains two integers $n$, $k$ ($1 \le n \le 2\cdot10^5$, $1 \le k \le 10^9$) — the number of questions and the number of possible answers to each question.

The following line contains $n$ integers $h_1, h_2, \dots, h_n$, ($1 \le h_{i} \le k)$ — answers to the questions.


-----Output-----

Output one integer: the number of answers suits satisfying the given condition, modulo $998\,244\,353$.


-----Examples-----
Input
3 3
1 3 1

Output
9

Input
5 5
1 1 4 2 2

Output
1000

Input
6 2
1 1 2 2 1 1

Output
16



-----Note-----

For the first example, valid answer suits are $[2,1,1], [2,1,2], [2,1,3], [3,1,1], [3,1,2], [3,1,3], [3,2,1], [3,2,2], [3,2,3]$.

Source: codeforces | Difficulty: VERY_HARD
Tags: Combinatorics

Mathematics


Solutions:
def main():
	M = 998244353
	(n, k, *h) = map(int, open(0).read().split())
	m = sum((i != j for (i, j) in zip(h, h[1:] + h[:1])))
	f = [0] * (m + 1)
	f[0] = b = 1
	for i in range(1, m + 1):
		f[i] = b = b * i % M
	inv = [0] * (m + 1)
	inv[m] = b = pow(f[m], M - 2, M)
	for i in range(m, 0, -1):
		inv[i - 1] = b = b * i % M
	comb = lambda n, k: f[n] * inv[n - k] * inv[k] % M
	print((pow(k, m, M) - sum((comb(m, i) * comb(m - i, i) * pow(k - 2, m - i - i, M) for i in range(m // 2 + 1)))) * pow(k, n - m, M) * pow(2, M - 2, M) % M)
main()


from bisect import *
from collections import *
from math import gcd, ceil, sqrt, floor, inf
from heapq import *
from itertools import *
from operator import add, mul, sub, xor, truediv, floordiv
from functools import *
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192

class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = 'x' in file.mode or 'r' not in file.mode
		self.write = self.buffer.write if self.writable else None

	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
		self.newlines = 0
		return self.buffer.read()

	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b'\n') + (not b)
			ptr = self.buffer.tell()
			(self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr))
		self.newlines -= 1
		return self.buffer.readline()

	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			(self.buffer.truncate(0), self.buffer.seek(0))

class IOWrapper(IOBase):

	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode('ascii'))
		self.read = lambda : self.buffer.read().decode('ascii')
		self.readline = lambda : self.buffer.readline().decode('ascii')
(sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout))
input = lambda : sys.stdin.readline().rstrip('\r\n')

def RL():
	return map(int, sys.stdin.readline().rstrip().split())

def RLL():
	return list(map(int, sys.stdin.readline().rstrip().split()))

def N():
	return int(input())
farr = [1]
ifa = []

def fact(x, mod=0):
	if mod:
		while x >= len(farr):
			farr.append(farr[-1] * len(farr) % mod)
	else:
		while x >= len(farr):
			farr.append(farr[-1] * len(farr))
	return farr[x]

def ifact(x, mod):
	global ifa
	ifa.append(pow(farr[-1], mod - 2, mod))
	for i in range(x, 0, -1):
		ifa.append(ifa[-1] * i % mod)
	ifa = ifa[::-1]

def per(i, j, mod=0):
	if i < j:
		return 0
	if not mod:
		return fact(i) // fact(i - j)
	return farr[i] * ifa[i - j] % mod

def com(i, j, mod=0):
	if i < j:
		return 0
	if not mod:
		return per(i, j) // fact(j)
	return per(i, j, mod) * ifa[j] % mod

def catalan(n):
	return com(2 * n, n) // (n + 1)

def linc(f, t, l, r):
	while l < r:
		mid = (l + r) // 2
		if t > f(mid):
			l = mid + 1
		else:
			r = mid
	return l

def rinc(f, t, l, r):
	while l < r:
		mid = (l + r + 1) // 2
		if t < f(mid):
			r = mid - 1
		else:
			l = mid
	return l

def ldec(f, t, l, r):
	while l < r:
		mid = (l + r) // 2
		if t < f(mid):
			l = mid + 1
		else:
			r = mid
	return l

def rdec(f, t, l, r):
	while l < r:
		mid = (l + r + 1) // 2
		if t > f(mid):
			r = mid - 1
		else:
			l = mid
	return l

def isprime(n):
	for i in range(2, int(n ** 0.5) + 1):
		if n % i == 0:
			return False
	return True

def binfun(x):
	c = 0
	for w in arr:
		c += ceil(w / x)
	return c

def lowbit(n):
	return n & -n

def inverse(a, m):
	a %= m
	if a <= 1:
		return a
	return (1 - inverse(m, a) * m) // a % m

class BIT:

	def __init__(self, arr):
		self.arr = arr
		self.n = len(arr) - 1

	def update(self, x, v):
		while x <= self.n:
			self.arr[x] += v
			x += x & -x

	def query(self, x):
		ans = 0
		while x:
			ans += self.arr[x]
			x &= x - 1
		return ans

class smt:

	def __init__(self, l, r, arr):
		self.l = l
		self.r = r
		self.value = (1 << 31) - 1 if l < r else arr[l]
		mid = (l + r) // 2
		if l < r:
			self.left = smt(l, mid, arr)
			self.right = smt(mid + 1, r, arr)
			self.value &= self.left.value & self.right.value

	def setvalue(self, x, val):
		if self.l == self.r:
			self.value = val
			return
		mid = (self.l + self.r) // 2
		if x <= mid:
			self.left.setvalue(x, val)
		else:
			self.right.setvalue(x, val)
		self.value = self.left.value & self.right.value

	def ask(self, l, r):
		if l <= self.l and r >= self.r:
			return self.value
		val = (1 << 31) - 1
		mid = (self.l + self.r) // 2
		if l <= mid:
			val &= self.left.ask(l, r)
		if r > mid:
			val &= self.right.ask(l, r)
		return val

class DSU:

	def __init__(self, n):
		self.c = [-1] * n

	def same(self, x, y):
		return self.find(x) == self.find(y)

	def find(self, x):
		if self.c[x] < 0:
			return x
		self.c[x] = self.find(self.c[x])
		return self.c[x]

	def union(self, u, v):
		(u, v) = (self.find(u), self.find(v))
		if u == v:
			return False
		if self.c[u] < self.c[v]:
			(u, v) = (v, u)
		self.c[u] += self.c[v]
		self.c[v] = u
		return True

	def size(self, x):
		return -self.c[self.find(x)]

class UFS:

	def __init__(self, n):
		self.parent = [i for i in range(n)]
		self.ranks = [0] * n

	def find(self, x):
		if x != self.parent[x]:
			self.parent[x] = self.find(self.parent[x])
		return self.parent[x]

	def union(self, u, v):
		(pu, pv) = (self.find(u), self.find(v))
		if pu == pv:
			return False
		if self.ranks[pu] >= self.ranks[pv]:
			self.parent[pv] = pu
			if self.ranks[pv] == self.ranks[pu]:
				self.ranks[pu] += 1
		else:
			self.parent[pu] = pv

def Prime(n):
	c = 0
	prime = []
	flag = [0] * (n + 1)
	for i in range(2, n + 1):
		if not flag[i]:
			prime.append(i)
			c += 1
		for j in range(c):
			if i * prime[j] > n:
				break
			flag[i * prime[j]] = prime[j]
			if i % prime[j] == 0:
				break
	return prime

def dij(s, graph):
	d = {}
	d[s] = 0
	heap = [(0, s)]
	seen = set()
	while heap:
		(dis, u) = heappop(heap)
		if u in seen:
			continue
		for v in graph[u]:
			if v not in d or d[v] > d[u] + graph[u][v]:
				d[v] = d[u] + graph[u][v]
				heappush(heap, (d[v], v))
	return d

def GP(it):
	return [(ch, len(list(g))) for (ch, g) in groupby(it)]

class DLN:

	def __init__(self, val):
		self.val = val
		self.pre = None
		self.next = None
t = 1
for i in range(t):
	mod = 998244353
	(n, k) = RL()
	h = RLL()
	res = GP(h)
	c = 0
	for (ch, cnt) in res:
		c += cnt - 1
	n = len(res)
	if res[-1][0] == res[0][0]:
		c += 1
		n -= 1
	ans = pow(k, c, mod)
	tmp = pow(k, n, mod)
	fact(n, mod)
	ifact(n, mod)
	p = [1]
	for i in range(n):
		p.append(p[-1] * (k - 2) % mod)
	for x in range(n // 2 + 1):
		tmp = (tmp - p[n - 2 * x] * fact(n, mod) % mod * ifa[x] % mod * ifa[x] % mod * ifa[n - 2 * x] % mod) % mod
	ans = ans * tmp % mod * pow(2, mod - 2, mod) % mod
	print(ans)


M = 998244353

class Factorial:

	def __init__(self, n):
		self.f = f = [0] * (n + 1)
		f[0] = b = 1
		for i in range(1, n + 1):
			f[i] = b = b * i % M
		self.inv = inv = [0] * (n + 1)
		inv[n] = b = pow(self.f[n], M - 2, M)
		for i in range(n, 0, -1):
			inv[i - 1] = b = b * i % M

	def factorial(self, i):
		return self.f[i]

	def ifactorial(self, i):
		return self.inv[i]

	def comb(self, n, k):
		if n >= k:
			return self.f[n] * self.inv[n - k] * self.inv[k] % M
		else:
			return 0

def main():
	(n, k, *h) = map(int, open(0).read().split())
	m = sum((i != j for (i, j) in zip(h, h[1:] + h[:1])))
	comb = Factorial(m).comb
	print((pow(k, m, M) - sum((comb(m, i) * comb(m - i, i) * pow(k - 2, m - i - i, M) for i in range(m // 2 + 1)))) * pow(k, n - m, M) * pow(2, M - 2, M) % M)
main()