#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
//using mint = modint1000000007;
//const int mod = 1000000007;
using mint = modint998244353;
const int mod = 998244353;
//const int INF = 1e9;
//const long long LINF = 1e18;
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define rep2(i,l,r)for(int i=(l);i<(r);++i)
#define rrep(i, n) for (int i = (n-1); i >= 0; --i)
#define rrep2(i,l,r)for(int i=(r-1);i>=(l);--i)
#define all(x) (x).begin(),(x).end()
#define allR(x) (x).rbegin(),(x).rend()
#define endl "\n"
#define P pair<int,int>
template<typename A, typename B> inline bool chmax(A & a, const B & b) { if (a < b) { a = b; return true; } return false; }
template<typename A, typename B> inline bool chmin(A & a, const B & b) { if (a > b) { a = b; return true; } return false; }
// combination mod prime
// https://www.youtube.com/watch?v=8uowVvQ_-Mo&feature=youtu.be&t=1619
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		assert(n < mod);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) { return com(n, k); }
	mint com(int n, int k) {
		//負の二項係数を考慮する場合にコメントアウトを外す
		//if (n < 0) return com(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
	mint inv(int n, int k) {
		//if (n < 0) return inv(-n, k) * (k % 2 ? -1 : 1);
		if (k < 0 || k > n) return 0;
		return ifact[n] * fact[k] * fact[n - k];
	}
	mint p(int n, int k) { return fact[n] * ifact[n - k]; }
}c(2000006);

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int n; cin >> n;
	vector<int>a(n);
	int sum = 0;
	rep(i, n) {
		cin >> a[i];
		sum += a[i];
	}
	mint ans = 0;
	rep(i, n) {
		mint sub = pow_mod(2, n - 1, mod);
		sub *= a[i];
		ans += sub;
	}
	if (sum >= 999630629) {
		int over = sum - 999630629;
		vector<mint>dp(over + 1);
		dp[0] = 1;
		sort(all(a));
		for (int i = 0; i < a.size();) {
			int j = i + 1;
			while ((j < n) && (a[i] == a[j]))j++;
			int cnt = j - i;
			rrep(j, over + 1 - a[i]) {
				if (0 != dp[i].val()) {
					for (int k = 1; (k <= cnt) && (j + a[i] * k <= over); ++k) {
						dp[j * a[i] * k] += dp[j] * c(cnt, k);
					}
				}
			}
			i = j;
		}
		rep(i, dp.size())ans -= dp[i] * 999630629;
	}
	cout << ans.val() << endl;
	return 0;
}