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 $[2,1,1], [2,1,2], [2,1,3], [3,1,1], [3,1,2], [3,1,3], [3,2,1], [3,2,2], [3,2,3]$.

Difficulty: interview

Starter code:

Solutions:
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()

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()

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()