Problem:
Score : 900 points

Problem Statement

Given are a permutation p_1, p_2, \dots, p_N of (1, 2, ..., N) and an integer K. Maroon performs the following operation for i = 1, 2, \dots, N - K + 1 in this order:

Shuffle p_i, p_{i + 1}, \dots, p_{i + K - 1} uniformly randomly.

Find the expected value of the inversion number of the sequence after all the operations are performed, and print it modulo 998244353.

More specifically, from the constraints of this problem, it can be proved that the expected value is always a rational number, which can be represented as an irreducible fraction \frac{P}{Q}, and that the integer R that satisfies R \times Q \equiv P \pmod{998244353}, 0 \leq R < 998244353 is uniquely determined. Print this R.

Here, the inversion number of a sequence a_1, a_2, \dots, a_N is defined to be the number of ordered pairs (i, j) that satisfy i < j, a_i > a_j.

Constraints

2 \leq N \leq 200,000

2 \leq K \leq N

(p_1, p_2, \dots, p_N) is a permutation of (1, 2, \dots, N).

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 p_2 ... p_N

Output

Print the expected value modulo 998244353.

Sample Input 1

3 2
1 2 3

Sample Output 1

1

The final sequence is one of (1, 2, 3), (2, 1, 3), (1, 3, 2), (2, 3, 1), each with probability \frac{1}{4}.
Their inversion numbers are 0, 1, 1, 2 respectively, so the expected value is 1.

Sample Input 2

10 3
1 8 4 9 2 3 7 10 5 6

Sample Output 2

164091855

Solution in C++ (GCC 9.2.1 with AC Library v1.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <atcoder/fenwicktree>
#include <atcoder/modint>
using namespace std;
using namespace atcoder;

using ll=long long;
#define rng(i,l,r) for(int i=int(l);i<int(r);i++)
#define rep(i,r) rng(i,0,r)
#define rrng(i,l,r) for(int i=int(r)-1;i>=int(l);i--)
#define rrep(i,r) rrng(i,0,r)
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define mt make_tuple
#define F first
#define S second
#define bg begin()
#define ed end()
#define all(x) x.bg,x.ed
#define si(x) int(x.size())
#define inf (INT_MAX/2-100)
#define infl (LLONG_MAX/3)
#ifdef LOCAL
#define dmp(x) cerr<<__LINE__<<' '<<#x<<' '<<x<<endl
#else
#define dmp(x) void(0)
#endif

template<class t,class u>bool chmax(t&a,u b){if(a<b)a=b;return a<b;}
template<class t,class u>bool chmin(t&a,u b){if(b<a)a=b;return b<a;}

template<class t>using vc=vector<t>;
template<class t>using vvc=vector<vector<t>>;

using pi=pair<int,int>;
using pl=pair<ll,ll>;
using vi=vc<int>;
using vl=vc<ll>;


ll readl(void){
	ll x;
	cin>>x;
	return x;
}
int readi(void){
	int x;
	cin>>x;
	return x;
}

string readstr(){
	string s;
	cin>>s;
	return s;
}
vi readvi(int n,int off=0){
	vi v(n);
	rep(i,n)v[i]=readi(),v[i]+=off;
	return v;
}

vl readvl(int n,int off=0){
	vl v(n);
	rep(i,n)v[i]=readl(),v[i]+=off;
	return v;
}

template<class t>
void print(t x,int suc=1){
	cout<<x;
	if(suc==1)cout<<"\n";
	if(suc==2)cout<<" ";
}

template<class t>
void print(const vc<t>&v,int suc=1){
	rep(i,si(v))print(v[i],i==int(si(v))-1?1:suc);
}

template<class t>
bool inc(t a,t b,t c){
	return !(c<b||b<a);
}

template<class t>
void compress(vc<t>&v){
	sort(all(v));
	v.erase(unique(all(v)),v.ed);
}

template<class t>
int lwb(const vc<t>&v,const t&a){
	return lower_bound(all(v),a)-v.bg;
}

template<class t>
struct Compress{
	vc<t>v;
	Compress()=default;
	Compress(const vc<t>&x){
		add(x);
	}
	Compress(const initializer_list<vc<t> >&x){
		for(auto &p:x)add(p);
	}
	void add(const t&x){
		v.eb(x);
	}
	void add(const vc<t>&x){
		copy(all(x),back_inserter(v));
	}
	void build(){
		compress(v);
	}
	int get(const t&x)const{
		return lwb(v,x);
	}
	vc<t>get(const vc<t>&x)const{
		vc<t>res(x);
		for(auto &p:res)p=get(p);
		return res;
	}
	const t &operator[](int x)const{
		return v[x];
	}
	int size(){
		return v.size();
	}
};
void Yes(bool ex=true){
	cout<<"Yes\n";
	if(ex)exit(0);
}
void YES(bool ex=true){
	cout<<"YES\n";
	if(ex)exit(0);
}
void No(bool ex=true){
	cout<<"No\n";
	if(ex)exit(0);
}
void NO(bool ex=true){
	cout<<"NO\n";
	if(ex)exit(0);
}
void orYes(bool x,bool ex=true){
	if(x)Yes(ex);
	else No(ex);
}
void orYES(bool x,bool ex=true){
	if(x)YES(ex);
	else NO(ex);
}
void Possible(bool ex=true){
	cout<<"Possible\n";
	if(ex)exit(0);
}
void POSSIBLE(bool ex=true){
	cout<<"POSSIBLE\n";
	if(ex)exit(0);
}
void Impossible(bool ex=true){
	cout<<"Impossible\n";
	if(ex)exit(0);
}
void IMPOSSIBLE(bool ex=true){
	cout<<"IMPOSSIBLE\n";
	if(ex)exit(0);
}
void orPossible(bool x,bool ex=true){
	if(x)Possible(ex);
	else Impossible(ex);
}
void orPOSSIBLE(bool x,bool ex=true){
	if(x)POSSIBLE(ex);
	else IMPOSSIBLE(ex);
}


using mint = modint;

ll n,k;
int main(void){
	cin.tie(0);
	ios::sync_with_stdio(0);
	mint::set_mod(998244353);
	cin>>n>>k;
	mint p=mint(k-1)/mint(k),ans=0,q=p.inv();
	fenwick_tree<mint>bit(n),bit2(n);
	rep(i,n){
		ll a,t=max(k-1,ll(i));
		cin>>a;
		a--;
		ans+=bit.sum(0,a)*p.pow(t)/(mint(2LL));
		ans-=bit.sum(a,n)*p.pow(t)/(mint(2LL));
		ans+=bit2.sum(a,n);
		bit.add(a,q.pow(t));
		bit2.add(a,mint(1LL));
	}
	print(ans.val());
}