#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>

using namespace std;

#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define REV(i,n) for(ll (i) = (n) - 1;(i) >= 0;--i)
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {REP(WW,n)cerr << v[WW] << ' ';cerr << endl;}
#define SHOW2d(v,WW,HH) {REP(W_,WW){REP(H_,HH)cerr << v[W_][H_] << ' ';cerr << endl;}}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define LLINF 1000000000000000000LL
#define MOD 998244353

typedef long long ll;
typedef pair<ll,ll> P;

/*
0-index
functionの中を書いて使う
segtree<ll> seg(n, [](ll a, ll b){return a + b;}, 0);
*/
template<typename T>
struct segtree{
	int n;
	vector<T> dat;
	T ZERO;
	function<T(T,T)> func;//要素と要素をマージする関数

	segtree(int n_, function<T(T,T)> func_, T zero):func(func_), ZERO(zero){
	  n = 1;
	  while(n < n_)n*=2;
	  dat.resize(2*n);
	  for(int i = 0;i < 2*n;i++)dat[i] = ZERO;
	}

	void update(int k, T a){
	  k += n - 1;dat[k] = a;
	  while(k > 0){
		k = (k-1)/2;
		dat[k] = func(dat[k*2+1],dat[k*2+2]);
	  }
	}

	void set(int k, T a) {
		k += n - 1;dat[k] = a;
	}

	void build(){
		for(int i = n - 2;i >= 0;i--){
			dat[i] = func(dat[2 * i + 1], dat[2 * i + 2]);
		}
	}

	T get(int k){return dat[k+n-1];}

	T que(int a,int b){
		T L = ZERO,R = ZERO;
		int A = a + n - 1;
		int B = b + n - 1;
		while(A < B){
			if((A & 1) == 0)L = func(L,dat[A++]);
            if((B & 1) == 0)R = func(dat[--B],R);
            A = A >> 1;
            B = B >> 1;
		}
		return func(L,R);
	}
};

//ａをbで割る
long long mod_div(long long a,long long b){
	long long tmp = MOD - 2,c = b,ret = 1;
	while(tmp > 0){
		if(tmp & 1){
			ret *= c;ret %= MOD;
		}
		c *= c;c %= MOD;tmp >>= 1;
	}
	return a*ret%MOD;
}

#define MAX_K 333333
vector<long long> fact(MAX_K);
long long combination(long long n, long long r){
    if(n < r || n < 0 || r < 0)
        return 0;
	if(fact[0] != 1){
		fact[0] = 1;
		for(long long i = 1;i < MAX_K;i++)fact[i] = (fact[i-1] * i) % MOD;
	}
    long long ret = fact[n];
	long long tmp = (fact[r] * fact[n-r]) % MOD;
    return mod_div(ret,tmp);
}

int main(){
	cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);

	ll n, k;cin >> n >> k;
	segtree<ll> seg_a(n + 2, [](ll a, ll b){return (a + b) % MOD;}, 0);
	segtree<ll> seg_b(n + 2, [](ll a, ll b){return (a + b) % MOD;}, 0);
	ll ans = 0;
	ll seica = mod_div(1, 2);
	ll base = 1;

	REP(i, n) {
		ll a; cin >> a;
		if(i >= k)base = base * mod_div(k - 1, k) % MOD;
		ans = (ans + seg_a.que(0, a) * base % MOD) % MOD;
		ans = (ans + seg_b.que(a + 1, n + 1) * 1 - seg_a.que(a + 1, n + 1) * base % MOD + MOD) % MOD;
		seg_a.update(a, seica * mod_div(1, base) % MOD);
		seg_b.update(a, 1);
		//REP(j, n+1)cout << seg_a.que(j, j+1) << " ";cout << endl;
	}

	cout << ans << endl;

	return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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.

Examples

Input

3 2
1 2 3


Output

1


Input

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


Output

164091855

### Response

```cpp
#include <iostream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <set>
#include <map>
#include <bitset>
#include <cmath>
#include <functional>

using namespace std;

#define REP(i,n) for(ll (i) = (0);(i) < (n);++i)
#define REV(i,n) for(ll (i) = (n) - 1;(i) >= 0;--i)
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define SHOW1d(v,n) {REP(WW,n)cerr << v[WW] << ' ';cerr << endl;}
#define SHOW2d(v,WW,HH) {REP(W_,WW){REP(H_,HH)cerr << v[W_][H_] << ' ';cerr << endl;}}
#define ALL(v) v.begin(),v.end()
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define LLINF 1000000000000000000LL
#define MOD 998244353

typedef long long ll;
typedef pair<ll,ll> P;

/*
0-index
functionの中を書いて使う
segtree<ll> seg(n, [](ll a, ll b){return a + b;}, 0);
*/
template<typename T>
struct segtree{
	int n;
	vector<T> dat;
	T ZERO;
	function<T(T,T)> func;//要素と要素をマージする関数

	segtree(int n_, function<T(T,T)> func_, T zero):func(func_), ZERO(zero){
	  n = 1;
	  while(n < n_)n*=2;
	  dat.resize(2*n);
	  for(int i = 0;i < 2*n;i++)dat[i] = ZERO;
	}

	void update(int k, T a){
	  k += n - 1;dat[k] = a;
	  while(k > 0){
		k = (k-1)/2;
		dat[k] = func(dat[k*2+1],dat[k*2+2]);
	  }
	}

	void set(int k, T a) {
		k += n - 1;dat[k] = a;
	}

	void build(){
		for(int i = n - 2;i >= 0;i--){
			dat[i] = func(dat[2 * i + 1], dat[2 * i + 2]);
		}
	}

	T get(int k){return dat[k+n-1];}

	T que(int a,int b){
		T L = ZERO,R = ZERO;
		int A = a + n - 1;
		int B = b + n - 1;
		while(A < B){
			if((A & 1) == 0)L = func(L,dat[A++]);
            if((B & 1) == 0)R = func(dat[--B],R);
            A = A >> 1;
            B = B >> 1;
		}
		return func(L,R);
	}
};

//ａをbで割る
long long mod_div(long long a,long long b){
	long long tmp = MOD - 2,c = b,ret = 1;
	while(tmp > 0){
		if(tmp & 1){
			ret *= c;ret %= MOD;
		}
		c *= c;c %= MOD;tmp >>= 1;
	}
	return a*ret%MOD;
}

#define MAX_K 333333
vector<long long> fact(MAX_K);
long long combination(long long n, long long r){
    if(n < r || n < 0 || r < 0)
        return 0;
	if(fact[0] != 1){
		fact[0] = 1;
		for(long long i = 1;i < MAX_K;i++)fact[i] = (fact[i-1] * i) % MOD;
	}
    long long ret = fact[n];
	long long tmp = (fact[r] * fact[n-r]) % MOD;
    return mod_div(ret,tmp);
}

int main(){
	cin.tie(0);cout.tie(0);ios::sync_with_stdio(false);

	ll n, k;cin >> n >> k;
	segtree<ll> seg_a(n + 2, [](ll a, ll b){return (a + b) % MOD;}, 0);
	segtree<ll> seg_b(n + 2, [](ll a, ll b){return (a + b) % MOD;}, 0);
	ll ans = 0;
	ll seica = mod_div(1, 2);
	ll base = 1;

	REP(i, n) {
		ll a; cin >> a;
		if(i >= k)base = base * mod_div(k - 1, k) % MOD;
		ans = (ans + seg_a.que(0, a) * base % MOD) % MOD;
		ans = (ans + seg_b.que(a + 1, n + 1) * 1 - seg_a.que(a + 1, n + 1) * base % MOD + MOD) % MOD;
		seg_a.update(a, seica * mod_div(1, base) % MOD);
		seg_b.update(a, 1);
		//REP(j, n+1)cout << seg_a.que(j, j+1) << " ";cout << endl;
	}

	cout << ans << endl;

	return 0;
}

```