#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll modc = 998244353;
class mint {
    ll x;
public:
    mint(ll x=0) : x((x%modc+modc)%modc) {}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += modc-a.x) >= modc) x -= modc;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= modc;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {return pow(modc-2);}
    mint& operator/=(const mint& a){ return (*this) *= a.inv();}
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }
    bool operator == (const mint& a) const{ return x == a.x;}
    bool operator != (const mint& a) const{ return x != a.x;}
    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
    friend istream& operator>>(istream& ip, mint &m) {
        ll t;
        ip >> t;
        m = mint(t);
        return ip;
    }
    ll val(){ return x;}
};

int main(){
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);

    ll N, z;
    cin >> N;
    vector<ll> A(N+1);
    mint ans=0, p, q;
    for (int i=1; i<=N; i++) cin >> A[i];

    //A_iの左の要素数x(x<=i-1)、右の要素数y(y<=N-i)
    //sum(x, y) (i-1)Cx * (N-i)Cy * (x+1)*(y+1)
    //sum(x=0 to z) zCx*(x+1)=2^{z-1}(z+2)=z*2^(z-1)+2^z

    for (int i=1; i<=N; i++){
        p = 0; q = 0;
        z = i-1;
        p += mint(2).pow(z);
        if (z >= 1) p += mint(2).pow(z-1) * z;
        z = N-i;
        q += mint(2).pow(z);
        if (z >= 1) q += mint(2).pow(z-1) * z;
        ans += p * q * A[i];
    }

    cout << ans << endl;

    return 0;
}