#include <bits/stdc++.h>
#include <atcoder/all>
// url 
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=b-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define pb(x) push_back(x);
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
typedef long long ll;
typedef long double lld;
using namespace std;
using namespace atcoder;
using mint = static_modint<998244353>;
// using mint = static_modint<1000000007>;
const ll mod=998244353;
//const ll mod=1e9+7;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
const string zton="0123456789";
const string atoz="abcdefghijklmnopqrstuvwxyz";
const string ATOZ="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ll inf=(1ll<<60);
// const int inf=(1<<30);
lld dist(lld x1,lld x2,lld y1,lld y2){
    lld res=(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
    res=sqrt(abs(res));
    return res;
}
lld arg(lld x,lld y){
    const lld eps=1e-8;
    lld res=0;
    if(abs(x)+abs(y)<=eps)return 0.0;
    else if(abs(x)<=eps){
        if(y>=0.0)return (M_PI/2);
        else return (M_PI/2+M_PI);
    }
    else if(abs(y)<=eps){
        if(x>=0.0)return 0.0;
        else return M_PI;
    }
    res=atan2(abs(y),abs(x));
    if(x<=0&&y>=0)res=(M_PI-res);
    else if(x<=0&&y<=0)res+=(M_PI);
    else if(x>=0&&y<=0)res=(M_PI*2-res);
    return res;
}
ll gcd(ll a,ll b){
    if(a==0||b==0)return a+b;
    ll r;
    r=a%b;
    if(r==0){
        return b;
    }
    else{
        return gcd(b,r);
    }
}
typedef pair<ll,int> P;

int main(void){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int N;cin>>N;
    P p[N];
    rep(i,0,N){
        cin>>p[i].first;
        p[i].second=i;
        p[i].first--;
    }
    sort(p,p+N);
    // reverse(p,p+N);
    fenwick_tree<mint> cnt(N),sum(N);
    mint power2[N+1];
    power2[0]=1;
    rep(i,1,N+1){
        power2[i]=power2[i-1]*2;
    }
    mint ans=0;
    rep(i,0,N){
        int idx=p[i].second;
        mint a=cnt.sum(idx,N);
        mint b=sum.sum(idx,N);
        a*=power2[N-1];
        b*=power2[idx];
        mint ad=a-b;
        ans+=ad;
        cnt.add(idx,1);
        sum.add(idx,power2[N-1-idx]);
        // cout << idx << endl;
        // cout << power2[N-1+idx].val() << endl;
        // cout << ad.val() << endl;
        // cout << a.val() << " " << b.val() << endl;

    }
    cout<<ans.val()<<endl;
}