#pragma GCC optimize("O3") #include #define ll long long #define rep(i,n) for(ll i=0;i<(n);i++) #define pll pair #define pii pair #define pq priority_queue #define pb push_back #define eb emplace_back #define fi first #define se second #define endl '\n' #define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0); #define lb(c,x) distance(c.begin(),lower_bound(all(c),x)) #define ub(c,x) distance(c.begin(),upper_bound(all(c),x)) using namespace std; inline int topbit(unsigned long long x){ return x?63-__builtin_clzll(x):-1; } inline int popcount(unsigned long long x){ return __builtin_popcountll(x); } inline int parity(unsigned long long x){//popcount%2 return __builtin_parity(x); } template inline bool chmax(T& a,T b){if(a inline bool chmin(T& a,T b){if(a>b){a=b;return 1;}return 0;} const ll INF=1e18; const ll mod = 998244353; struct mint { ll x; // typedef long long ll; mint(ll x=0):x((x%mod+mod)%mod){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint a) { if ((x += mod-a.x) >= mod) x -= mod; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;} mint operator+(const mint a) const { return mint(*this) += a;} mint operator-(const mint a) const { return mint(*this) -= a;} mint operator*(const mint a) const { return mint(*this) *= a;} mint pow(ll t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime mod mint inv() const { return pow(mod-2);} mint& operator/=(const mint a) { return *this *= a.inv();} mint operator/(const mint a) const { return mint(*this) /= a;} }; istream& operator>>(istream& is, mint& a) { return is >> a.x;} ostream& operator<<(ostream& os, const mint& a) { return os << a.x;} // combination mod prime // https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619 struct combination { vector fact, ifact; combination(ll n):fact(n+1),ifact(n+1) { assert(n < mod); fact[0] = 1; for (ll i = 1; i <= n; ++i) fact[i] = fact[i-1]*i; ifact[n] = fact[n].inv(); for (ll i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i; } mint operator()(ll n, ll k) { if (k < 0 || k > n) return 0; return fact[n]*ifact[k]*ifact[n-k]; } mint p(ll n, ll k) { return fact[n]*ifact[n-k]; } } c(1000005); template struct BIT { int n; vector d; BIT(int n=0):n(n),d(n+1) {} void add(int i, T x=1) { for (i++; i <= n; i += i&-i) { //「i += i&-i」は区間を表すindexに最下位ビットを足したもの //「-i = ~i-1」が成り立つ d[i] += x; } //更新 } T sum(int i) { T x = 0; for (i++; i; i -= i&-i) { x += d[i]; } //iが0になったら終わり //取得 return x; } }; int main(){ ll n; cin >> n; vector a(n),p(n),q(n); vector b(n); BIT s(n+1),v(n+1); BIT t(n+1),u(n+1); rep(i,n){ cin >> a[i]; b[i].fi=a[i]; b[i].se=i; } sort(b.begin(),b.end()); mint ans=0; rep(i,n){ mint cnt=s.sum(n)-s.sum(b[i].se); mint sum1=t.sum(n)-t.sum(b[i].se); mint sum2=u.sum(n)-u.sum(b[i].se); mint cnt2=v.sum(n)-v.sum(b[i].se); s.add(b[i].se,1); t.add(b[i].se,b[i].fi); u.add(b[i].se,sum1+cnt*b[i].fi); v.add(b[i].se,cnt); ans+=sum2+cnt2*b[i].fi; } cout << ans << endl; return 0; }