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

using Int = long long;
const char newl = '\n';

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
template<typename T> void drop(const T &x){cout<<x<<endl;exit(0);}
template<typename T=Int>
vector<T> read(size_t n){
  vector<T> ts(n);
  for(size_t i=0;i<n;i++) cin>>ts[i];
  return ts;
}


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;}
  Mint operator-(Mint a) const{return Mint(v)-=a;}
  Mint operator*(Mint a) const{return Mint(v)*=a;}
  Mint operator/(Mint a) const{return Mint(v)/=a;}

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static Mint comb(long long n,Int k){
    Mint num(1),dom(1);
    for(Int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}


template<typename T>
class BIT{
private:
  // \sum_{j < i}  v[j]
  T sum(Int i){
    T s(0);
    for(Int x=i;x>0;x-=(x&-x))
      s+=bit[x];
    return s;
  }
public:
  Int n;
  vector<T> bit;
  BIT(Int n_):n(n_+1),bit(n+1,0){}

  // v[i] += a
  void add(Int i,T a){
    for(Int x=++i;x<=n;x+=(x&-x))
      bit[x]+=a;
  }
  // \sum_{l <= i < r} v[i]
  T query(Int l,Int r){return sum(r)-sum(l);}

  // min({x | sum(x) >= w})
  Int lower_bound(const T w){
    if(w<=0) return 0;
    T r=w;
    Int x=0,p=1;
    while(p<n) p<<=1;
    for(Int k=p;k>0;k>>=1){
      if(x+k<=n&&bit[x+k]<r){
        r-=bit[x+k];
        x+=k;
      }
    }
    x++;
    assert(sum(x-1)<w and sum(x)>=w);
    return x;
  }

  // min({x | sum(x) > w})
  Int upper_bound(T w){return lower_bound(w+1);}
};

//INSERT ABOVE HERE
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);
  Int n,k;
  cin>>n>>k;
  auto ps=read(n);
  // for(Int &p:ps) p--;

  using M = Mint<Int, 998244353>;

  M ans=M(k*(k-1)/2)/M(2);

  BIT<M> bit(n+10),cnt(n+10);
  for(Int i=0;i<k;i++){
    Int p=ps[i];
    bit.add(p,M(1));
    cnt.add(p,M(1));
  }

  // assert(n!=k);

  M ofs{1};
  M ratio=M(k-1)/M(k);
  for(Int i=k;i<n;i++){
    ofs*=ratio;
    Int p=ps[i];
    ans+=(bit.query(0,n+10)*ofs)/M(2);
    ans+=(cnt.query(p,n+10)-bit.query(p,n+10)*ofs);
    bit.add(p,M(1)/ofs);
    cnt.add(p,M(1));
  }

  cout<<ans<<newl;
  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 <bits/stdc++.h>
using namespace std;

using Int = long long;
const char newl = '\n';

template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
template<typename T> void drop(const T &x){cout<<x<<endl;exit(0);}
template<typename T=Int>
vector<T> read(size_t n){
  vector<T> ts(n);
  for(size_t i=0;i<n;i++) cin>>ts[i];
  return ts;
}


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;}
  Mint operator-(Mint a) const{return Mint(v)-=a;}
  Mint operator*(Mint a) const{return Mint(v)*=a;}
  Mint operator/(Mint a) const{return Mint(v)/=a;}

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  static Mint comb(long long n,Int k){
    Mint num(1),dom(1);
    for(Int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}


template<typename T>
class BIT{
private:
  // \sum_{j < i}  v[j]
  T sum(Int i){
    T s(0);
    for(Int x=i;x>0;x-=(x&-x))
      s+=bit[x];
    return s;
  }
public:
  Int n;
  vector<T> bit;
  BIT(Int n_):n(n_+1),bit(n+1,0){}

  // v[i] += a
  void add(Int i,T a){
    for(Int x=++i;x<=n;x+=(x&-x))
      bit[x]+=a;
  }
  // \sum_{l <= i < r} v[i]
  T query(Int l,Int r){return sum(r)-sum(l);}

  // min({x | sum(x) >= w})
  Int lower_bound(const T w){
    if(w<=0) return 0;
    T r=w;
    Int x=0,p=1;
    while(p<n) p<<=1;
    for(Int k=p;k>0;k>>=1){
      if(x+k<=n&&bit[x+k]<r){
        r-=bit[x+k];
        x+=k;
      }
    }
    x++;
    assert(sum(x-1)<w and sum(x)>=w);
    return x;
  }

  // min({x | sum(x) > w})
  Int upper_bound(T w){return lower_bound(w+1);}
};

//INSERT ABOVE HERE
signed main(){
  cin.tie(0);
  ios::sync_with_stdio(0);
  Int n,k;
  cin>>n>>k;
  auto ps=read(n);
  // for(Int &p:ps) p--;

  using M = Mint<Int, 998244353>;

  M ans=M(k*(k-1)/2)/M(2);

  BIT<M> bit(n+10),cnt(n+10);
  for(Int i=0;i<k;i++){
    Int p=ps[i];
    bit.add(p,M(1));
    cnt.add(p,M(1));
  }

  // assert(n!=k);

  M ofs{1};
  M ratio=M(k-1)/M(k);
  for(Int i=k;i<n;i++){
    ofs*=ratio;
    Int p=ps[i];
    ans+=(bit.query(0,n+10)*ofs)/M(2);
    ans+=(cnt.query(p,n+10)-bit.query(p,n+10)*ofs);
    bit.add(p,M(1)/ofs);
    cnt.add(p,M(1));
  }

  cout<<ans<<newl;
  return 0;
}

```