#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll=long long;
using ld=long double;
ld pie=3.14159265359;
ll inf=10000000000000001;
ll mod=998244353;
template <typename T>
vector<T> compress(vector<T> &X) {
    // ソートした結果を vals に
    vector<T> vals = X;
    sort(vals.begin(), vals.end());
    // 隣り合う重複を削除(unique), 末端のゴミを削除(erase)
    vals.erase(unique(vals.begin(), vals.end()), vals.end());
    // 各要素ごとに二分探索で位置を求める
    for (int i = 0; i < (int)X.size(); i++) {
        X[i] = lower_bound(vals.begin(), vals.end(), X[i]) - vals.begin();
    }
    return vals;
}
ll op(ll a,ll b){
    return (a+b)%mod;
}
ll e(){
    return 0;
}
int main(){
    ll n;
    cin >> n;
    vector<ll>a(n);
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    vector<ll>aa=a;
    vector<ll>b=compress(aa);
    map<ll,ll>memo;
    for (ll i = 0; i < b.size(); i++)
    {
        memo[b[i]]=i;
    }
    vector<ll>c(b.size(),0),d(b.size(),0),ee(b.size(),0),f(b.size(),0);
    for (ll i = 0; i < a.size(); i++)
    {
        c[memo[a[i]]]+=1;
        ee[memo[a[i]]]+=a[i];
        ee[memo[a[i]]]%=mod;
    }
    ll last=b.size();
    segtree<ll,op,e>seg1(c),seg2(d),seg3(ee),seg4(f);
    ll ans=0;
    for (ll i = 0; i < n; i++)
    {
        ll y=memo[a[i]];
        ll syoko=seg1.prod(0,y);
        ll syo=seg3.prod(0,y);
        ll daiko=seg2.prod(y+1,last);
        ll dai=seg4.prod(y+1,last);
        ans+=syoko*dai;
        ans%=mod;
        ans+=daiko*syo;
        ans%=mod;
        ans+=((syoko*daiko)%mod)*a[i];
        seg1.set(y,seg1.get(y)-1);
        seg3.set(y,seg3.get(y)-a[i]);
        seg2.set(y,seg2.get(y)+1);
        seg4.set(y,seg4.get(y)+a[i]);
    }
    cout << ans << endl;
}