#include #include using namespace std; using namespace atcoder; typedef long long int ll; typedef long double ld; typedef vector vi; typedef vector vl; typedef vector vvl; typedef vector vvvl; typedef vector vvvvl; typedef vector vb; typedef vector vvb; typedef vector vvvb; typedef vector vvvvb; typedef pair pl; typedef pair ppl; typedef pair pppl; typedef pair pppppl; #define rep(i,a,b) for(int i=(a);i<(b);i++) #define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--) #define all(a) begin(a),end(a) #define sz(a) (int)(a).size() #define F first #define S second #define bs(A,x) binary_search(all(A),x) #define lb(A,x) (ll)(lower_bound(all(A),x)-A.begin()) #define ub(A,x) (ll)(upper_bound(all(A),x)-A.begin()) #define cou(A,x) (ll)(upper_bound(all(A),x)-lower_bound(all(A),x)) templateusing min_priority_queue=priority_queue,greater>; templatebool chmax(T&a,T b){if(abool chmin(T&a,T b){if(b vm; typedef vector vvm; typedef vector vvvm; typedef vector vvvvm; ostream&operator<<(ostream&os,mint a){os<>(istream&is,mint&a){int x;is>>x;a=mint(x);return is;} //*/ templateostream&operator<<(ostream&os,pairp){os<istream&operator>>(istream&is,pair&p){is>>p.F>>p.S;return is;} templateostream&operator<<(ostream&os,vectorv){rep(i,0,sz(v))os<istream&operator>>(istream&is,vector&v){for(T&in:v)is>>in;return is;} ll f(vectorQ){ vectorT; for(auto[x,y]:Q){ if(sz(T)>1&&T[sz(T)-2].first==x&&T[sz(T)-1].first==3){ T[sz(T)-2].second+=T[sz(T)-1].second+y; T.erase(T.end()-1); } else T.emplace_back(pl(x,y)); } ll N=sz(T); vl DP(N+1); DP[1]=T[0].second*T[0].second; rep(i,2,N+1)DP[i]=max(DP[i-1]+T[i-1].second*T[i-1].second,DP[i-2]+(T[i-1].second+T[i-2].second)*(T[i-1].second+T[i-2].second)); return DP[N]; } int main(){ cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit); ll N;cin>>N; vl A(N);cin>>A; ll M;cin>>M; vl B(M);cin>>B; vl C(N+1),D(M+1); rep(i,0,N)C[i+1]=A[i]+C[i]; rep(i,0,M)D[i+1]=B[i]+D[i]; vl X=C; X.insert(X.end(),all(D)); sort(all(X)); X.erase(unique(all(X)),X.end()); ll L=sz(X)-1; vl S(L); ll t=0; rep(i,0,N)while(X[t]Q; rep(k,j,i)Q.emplace_back(pl(S[k],X[k+1]-X[k])); ans+=f(Q); i--; } cout<