#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define FORR2(x,y,arr) for(auto& [x,y]:arr)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
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(a>b){a=b;return 1;}return 0;}
//-------------------------------------------------------

ll N;
ll A[202020];
int C[505050];
const ll mo=998244353;
const ll P=999630629;
ll dp[505050];

ll modpow(ll a, ll n = mo-2) {
	ll r=1;a%=mo;
	while(n) r=r*((n%2)?a:1)%mo,a=a*a%mo,n>>=1;
	return r;
}

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N;
	ll sum=0;
	ll ret=0;
	FOR(i,N) {
		cin>>A[i];
		sum+=A[i];
		C[A[i]]++;
	}
	ret=sum*modpow(2,N-1)%mo;
	if(sum>=P) {
		unordered_map<ll,ll> dp;
		dp[sum-P]=1;
		sort(A,A+N);
		reverse(A,A+N);
		
		FOR(i,N) {
			vector<pair<ll,ll>> add;
			FORR2(a,b,dp) if(a>=A[i]) add.push_back({a-A[i],b});
			FORR2(a,b,add) dp[a]=(dp[a]+b)%mo;
		}
		FOR(j,sum-P+1) ret-=dp[j]*P%mo;
	}
	cout<<(ret%mo+mo)%mo<<endl;
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}