#include <iostream>
#include <algorithm>
#include <vector>
#include <atcoder/modint>
#define rep(i, n) for(i = 0; i < n; i++)
#define int long long
using namespace std;
using namespace atcoder;
using mint = modint998244353;

int gcd(int a, int b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

void experiment() {
	int n, i, j, k, l;
	int mats[16][16];
	int cnts[16];
	for (n = 1; n <= 15; n++) {
		vector<int> kiyo(n);
		int cnt = 0;
		for (i = 1; i < (1 << n); i++) {
			vector<int> vec;
			rep(j, n) if ((i >> j) % 2) vec.push_back(j);
			rep(j, vec.size()) {
				for (k = j; k < vec.size(); k++) {
					cnt++;
					for (l = j; l <= k; l++) kiyo[vec[l]]++;
				}
			}
		}
		rep(i, n) mats[n][i] = kiyo[i];
		cnts[n] = cnt;
	}

	for (n = 2; n <= 15; n++) {
		rep(i, n - 1) {
			int si = mats[n][i];
			int bo = mats[n - 1][i];
			int g = gcd(si, bo);
			si /= g; bo /= g;
			printf("%5lld/%5lld ", si, bo);
		}
		cout << endl;
	}
}

//長さiのときの先頭要素の寄与a[i]を求める
vector<mint> make_seqs(int n) {
	int i;
	vector<mint> a(n);
	a[1] = 1;
	for (i = 2; i < n; i++) {
		if (i % 2 == 0) {
			a[i] = a[i - 1] * mint(i + 1) / mint(i / 2);	
		}
		else {
			a[i] = a[i - 1] * mint(2 * i + 2) / mint(i);
		}
	}
	return a;
}

signed main() {
	int n, i;
	cin >> n;
	vector<mint> pekora = make_seqs(n + 1);
	mint ans = 0;
	rep(i, n) {
		int x; cin >> x;
		ans += mint(x) * pekora[i + 1] * pekora[n - i];
	}
	cout << ans.val() << endl;
	return 0;
}