#include <bits/stdc++.h>
    using namespace std;
    #include <math.h>
    #include <iomanip>
    #include <cstdint>
    #include <string>
    #include <sstream>
    template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
    template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
    #define rep(i,n) for (int i = 0; i < (n); ++i)
    typedef long long ll;
    typedef unsigned long long ull;
    using P=pair<ll,ll>;
    const ll INF=1001001001;
    const int mod=998244353;
    
    struct mint {
      ll x; 
      mint(ll x=0):x((x%mod+mod)%mod){}
      mint operator-() const { return mint(-x);}
      mint& operator+=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
      }
      mint& operator-=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
      }
      mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
      mint operator+(const mint a) const { return mint(*this) += a;}
      mint operator-(const mint a) const { return mint(*this) -= a;}
      mint operator*(const mint a) const { return mint(*this) *= 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(mod-2);}
      mint& operator/=(const mint a) { return *this *= a.inv();}
      mint operator/(const mint a) const { return mint(*this) /= a;}
     };
      istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
      ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
      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) {
        if (k < 0 || k > n) return 0;
        return fact[n]*ifact[k]*ifact[n-k];
      }
     } c(200005);
    
    const int p=10000;

    mint dp[105][2*p+5];
    
    void solve(){
        int n;
        cin>>n;
        vector<int>a(n);
        rep(i,n) cin>>a[i];
        int p=10000;
        
        dp[0][p]=1;
        rep(i,n){
            rep(j,2*p+1){
                if(j+a[i]>2*p||j-a[i]<0){continue;}
                if(dp[i][j].x==0){continue;}
                dp[i+1][j+a[i]]+=dp[i][j];
                dp[i+1][j-a[i]]+=dp[i][j];
            }
        }
        mint ans=0;
        rep(i,2*p+1){
            if(dp[n][i].x!=0){
                if(i<=p){ans+=dp[n][i]*(p-i);}
                else{ans+=dp[n][i]*(i-p);}
            }
        }
        cout<<ans<<endl;
        
    }
 
    int main(){
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        
        solve();  

        return 0;
    }