#include #pragma GCC optimize("Ofast") #define _GLIBCXX_DEBUG using namespace std; using std::cout; using std::cin; using std::endl; using ll=long long; using ld=long double; ll ILL=2167167167167167167; const int INF=2100000000; const ll mod=998244353; #define rep(i,a) for (ll i=0;i using _pq = priority_queue, greater>; template ll LB(vector &v,T a){return lower_bound(v.begin(),v.end(),a)-v.begin();} template ll UB(vector &v,T a){return upper_bound(v.begin(),v.end(),a)-v.begin();} template bool chmin(T &a,const T &b){if(a>b){a=b;return 1;}else return 0;} template bool chmax(T &a,const T &b){if(a void So(vector &v) {sort(v.begin(),v.end());} template void Sore(vector &v) {sort(v.begin(),v.end(),[](T x,T y){return x>y;});} void yneos(bool a){if(a) cout<<"Yes\n"; else cout<<"No\n";} template void vec_out(vector &p){for(int i=0;i<(int)(p.size());i++){if(i) cout<<" ";cout< struct seg_sum{ int n; int seg_size; std::vector seg; T e; //要素数、単位元 typeは+演算が定義されているもの seg_sum(int k,T e_in){ e=e_in; n=k; seg_size=1; while(seg_size //p,qのなかみは、重複ありで内容が同じならおけ long long inversion(std::vector &p,std::vector &q){ int n=p.size(); assert(p.size()==q.size()); po167::seg_sum S(n,0); std::vector p_index(n,-1); std::map> m; for(int i=0;i //qのなかみは、0~nの順列の並び替えである必要がある long long inversion(std::vector &q){ int n=q.size(); po167::seg_sum S(n,0); std::vector p_index(n); for(int i=0;i fact; std::vector rev; std::vector fact_rev; combination(int max,long long mod):upper(max),MOD(mod),fact(max+1),rev(max+1),fact_rev(max+1){ for(long long i=0;i<=max;i++){ if(i<2){ fact[i]=1; fact_rev[i]=1; rev[i]=1; continue; } fact[i]=(fact[i-1]*i)%mod; rev[i]=mod-((mod/i)*rev[mod%i])%mod; fact_rev[i]=(fact_rev[i-1]*rev[i])%mod; } } long long Comb(int x,int y){ assert(upper>=x); if (x0){ a*=2; if(H%a!=0){ H-=a/2; c*=b; c%=z; } b*=b; b%=z; } //ここまで return c; } void solve(); // oddloop int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int t=1; //cin>>t; rep(i,t) solve(); } void solve(){ ll N; cin>>N; vector A(N); ll X=0; map m; rep(i,N){ cin>>A[i]; X+=i-m[A[i]]; m[A[i]]++; } if(N==1){ cout<<"0\n"; return ; } auto B=A; So(B); ll I=inversion(A,B); ll ans=N+1; ans=(ans*jyo(3,mod-2,mod))%mod; ans=(ans*I)%mod; combination table(N,mod); ll S=0; rep(i,N) S=(S+i+1)%mod; S=(S*S)%mod; rep(i,N) S=(S-(i+1)*(i+1))%mod; S=(S*jyo(2ll*N*N,mod-2,mod))%mod; S=(S*(X%mod))%mod; ans=(ans+S+mod)%mod; rep(i,N){ ans=(ans*table.Comb(N,i+1))%mod; } cout<