#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <algorithm>
#include <cassert>
#include <random>
#include <iomanip>
#include <bitset>
#include <array>
#define FOR(i, n, m) for(ll i = (n); i < (ll)(m); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define pb push_back
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 998244353;
constexpr long double eps = 1e-6;
 
template<typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
    os << to_string(p.first) << " " << to_string(p.second);
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
    REP(i, v.size()) {
        if(i) os << " ";
        os << v[i];
    }
    return os;
}
 
struct modint {
    ll n;
public:
    modint(const ll n = 0) : n((n % mod + mod) % mod) {}
    static modint pow(modint a, int m) {
        modint r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; }
            a = (a * a); m /= 2;
        }
        return r;
    }
    modint &operator++() { *this += 1; return *this; }
    modint &operator--() { *this -= 1; return *this; }
    modint operator++(int) { modint ret = *this; *this += 1; return ret; }
    modint operator--(int) { modint ret = *this; *this -= 1; return ret; }
    modint operator~() const { return (this -> pow(n, mod - 2)); } // inverse
    friend bool operator==(const modint& lhs, const modint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const modint& lhs, const modint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const modint& lhs, const modint& rhs) {
        return lhs.n > rhs.n;
    }
    friend modint &operator+=(modint& lhs, const modint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend modint &operator-=(modint& lhs, const modint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend modint &operator*=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend modint &operator/=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend modint operator+(const modint& lhs, const modint& rhs) {
        return modint(lhs.n + rhs.n);
    }
    friend modint operator-(const modint& lhs, const modint& rhs) {
        return modint(lhs.n - rhs.n);
    }
    friend modint operator*(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * rhs.n);
    }
    friend modint operator/(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, modint m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, modint m) { os << m.n; return os; }
 
#define MAX_N 1010101
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
long long mod_inverse(long long a, long long m) {
    long long x, y;
    if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
    else return -1;
}
vector<long long> fact(MAX_N+1, inf);
long long mod_fact(long long n, long long& e) {
    if(fact[0] == inf) {
        fact[0]=1;
        if(MAX_N != 0) fact[1]=1;
        for(ll i = 2; i <= MAX_N; ++i) {
            fact[i] = (fact[i-1] * i) % mod;
        }
    }
    e = 0;
    if(n == 0) return 1;
    long long res = mod_fact(n / mod, e);
    e += n / mod;
    if((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
    return (res * fact[n % mod]) % mod;
}
// return nCk
long long mod_comb(long long n, long long k) {
    if(n < 0 || k < 0 || n < k) return 0;
    long long e1, e2, e3;
    long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
    if(e1 > e2 + e3) return 0;
    return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
 
using mi = modint;
 
mi mod_pow(mi a, ll n) {
    mi ret = 1;
    mi tmp = a;
    while(n > 0) {
        if(n % 2) ret *= tmp;
        tmp = tmp * tmp;
        n /= 2;
    }
    return ret;
}
 
ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll N = 202020;
mi bit[202020];
void add(ll a, mi w) {
    a++;
    for (ll x = a; x <= N; x += x & -x) bit[x] += w;
}
mi sum(ll a) {
    a++;
    mi ret = 0;
    for (ll x = a; x > 0; x -= x & -x) ret += bit[x];
    return ret;
}
// return the sum of [a, b)
mi sum(ll a, ll b) {
    mi ret = 0;
    if(b == 0) return 0;
    ret += sum(b - 1);
    if(a) ret -= sum(a - 1);
    return ret;
}

mi bit2[202020];
void add2(ll a, mi w) {
    a++;
    for (ll x = a; x <= N; x += x & -x) bit2[x] += w;
}
mi sum2(ll a) {
    a++;
    mi ret = 0;
    for (ll x = a; x > 0; x -= x & -x) ret += bit2[x];
    return ret;
}
// return the sum of [a, b)
mi sum2(ll a, ll b) {
    mi ret = 0;
    if(b == 0) return 0;
    ret += sum2(b - 1);
    if(a) ret -= sum2(a - 1);
    return ret;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    vector<int> p(n);
    REP(i, n) cin >> p[i];
    mi val = (k - 1); val /= k;
    mi div2 = mod_pow(2, mod - 2);
    mi ans = 0;
    
    mi val2 = mod_pow(val, k - 1);
    
    mi val3 = mod_pow(val, mod - 2);
    mi val4 = mod_pow(val3, k - 1);
    
    REP(i, n) {
        int lo = sum(0, p[i]).n, hi = sum(p[i], n + 3).n;
        
        if(i >= k) {
            val2 *= val;
            val4 *= val3;
        }
        
        ans += div2 * val2 * sum2(0, p[i]);
        ans += hi;
        ans -= div2 * val2 * sum2(p[i], n + 3);
        
        add(p[i], 1);
        add2(p[i], val4);
    }
    
    cout << ans << endl;
    return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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 <vector>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <algorithm>
#include <cassert>
#include <random>
#include <iomanip>
#include <bitset>
#include <array>
#define FOR(i, n, m) for(ll i = (n); i < (ll)(m); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define pb push_back
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 998244353;
constexpr long double eps = 1e-6;
 
template<typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
    os << to_string(p.first) << " " << to_string(p.second);
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
    REP(i, v.size()) {
        if(i) os << " ";
        os << v[i];
    }
    return os;
}
 
struct modint {
    ll n;
public:
    modint(const ll n = 0) : n((n % mod + mod) % mod) {}
    static modint pow(modint a, int m) {
        modint r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; }
            a = (a * a); m /= 2;
        }
        return r;
    }
    modint &operator++() { *this += 1; return *this; }
    modint &operator--() { *this -= 1; return *this; }
    modint operator++(int) { modint ret = *this; *this += 1; return ret; }
    modint operator--(int) { modint ret = *this; *this -= 1; return ret; }
    modint operator~() const { return (this -> pow(n, mod - 2)); } // inverse
    friend bool operator==(const modint& lhs, const modint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const modint& lhs, const modint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const modint& lhs, const modint& rhs) {
        return lhs.n > rhs.n;
    }
    friend modint &operator+=(modint& lhs, const modint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend modint &operator-=(modint& lhs, const modint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend modint &operator*=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend modint &operator/=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend modint operator+(const modint& lhs, const modint& rhs) {
        return modint(lhs.n + rhs.n);
    }
    friend modint operator-(const modint& lhs, const modint& rhs) {
        return modint(lhs.n - rhs.n);
    }
    friend modint operator*(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * rhs.n);
    }
    friend modint operator/(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, modint m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, modint m) { os << m.n; return os; }
 
#define MAX_N 1010101
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
long long mod_inverse(long long a, long long m) {
    long long x, y;
    if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
    else return -1;
}
vector<long long> fact(MAX_N+1, inf);
long long mod_fact(long long n, long long& e) {
    if(fact[0] == inf) {
        fact[0]=1;
        if(MAX_N != 0) fact[1]=1;
        for(ll i = 2; i <= MAX_N; ++i) {
            fact[i] = (fact[i-1] * i) % mod;
        }
    }
    e = 0;
    if(n == 0) return 1;
    long long res = mod_fact(n / mod, e);
    e += n / mod;
    if((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
    return (res * fact[n % mod]) % mod;
}
// return nCk
long long mod_comb(long long n, long long k) {
    if(n < 0 || k < 0 || n < k) return 0;
    long long e1, e2, e3;
    long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
    if(e1 > e2 + e3) return 0;
    return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}
 
using mi = modint;
 
mi mod_pow(mi a, ll n) {
    mi ret = 1;
    mi tmp = a;
    while(n > 0) {
        if(n % 2) ret *= tmp;
        tmp = tmp * tmp;
        n /= 2;
    }
    return ret;
}
 
ll gcd(ll a, ll b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll N = 202020;
mi bit[202020];
void add(ll a, mi w) {
    a++;
    for (ll x = a; x <= N; x += x & -x) bit[x] += w;
}
mi sum(ll a) {
    a++;
    mi ret = 0;
    for (ll x = a; x > 0; x -= x & -x) ret += bit[x];
    return ret;
}
// return the sum of [a, b)
mi sum(ll a, ll b) {
    mi ret = 0;
    if(b == 0) return 0;
    ret += sum(b - 1);
    if(a) ret -= sum(a - 1);
    return ret;
}

mi bit2[202020];
void add2(ll a, mi w) {
    a++;
    for (ll x = a; x <= N; x += x & -x) bit2[x] += w;
}
mi sum2(ll a) {
    a++;
    mi ret = 0;
    for (ll x = a; x > 0; x -= x & -x) ret += bit2[x];
    return ret;
}
// return the sum of [a, b)
mi sum2(ll a, ll b) {
    mi ret = 0;
    if(b == 0) return 0;
    ret += sum2(b - 1);
    if(a) ret -= sum2(a - 1);
    return ret;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    vector<int> p(n);
    REP(i, n) cin >> p[i];
    mi val = (k - 1); val /= k;
    mi div2 = mod_pow(2, mod - 2);
    mi ans = 0;
    
    mi val2 = mod_pow(val, k - 1);
    
    mi val3 = mod_pow(val, mod - 2);
    mi val4 = mod_pow(val3, k - 1);
    
    REP(i, n) {
        int lo = sum(0, p[i]).n, hi = sum(p[i], n + 3).n;
        
        if(i >= k) {
            val2 *= val;
            val4 *= val3;
        }
        
        ans += div2 * val2 * sum2(0, p[i]);
        ans += hi;
        ans -= div2 * val2 * sum2(p[i], n + 3);
        
        add(p[i], 1);
        add2(p[i], val4);
    }
    
    cout << ans << endl;
    return 0;
}

```