#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll N, Q, Sum, C1, C[202020], R[202020];
vector<pair<ll,ll>> G[202020];
void TreeDP(int v, int b=-1, ll up=0, ll dw=0){
for(auto [i,w] : G[v]) if(i == b) C1 += w, up += w;
C[v] = dw - up;
for(auto [i,w] : G[v]) if(i != b) TreeDP(i, v, up, dw+w);
}
ll CostToRoot(int root){
return C1 + C[root];
}
vector<pair<ll,ll>> PathsFromRoot(int root){
vector<pair<ll,ll>> paths;
function<ll(int,int,int)> dfs = [&](int st, int v, int b) -> ll {
ll mx = 0;
for(auto [i,w] : G[v]){
if(i == b) continue;
ll nxt = dfs(st, i, v) + w;
if(nxt > mx) swap(mx, nxt);
if(mx != 0) paths.emplace_back(nxt, st);
}
return mx;
};
for(auto [i,w] : G[root]) paths.emplace_back(dfs(i, i, root) + w, i);
return paths;
}
void Go(int root){
ll to_root = CostToRoot(root);
vector<pair<ll,ll>> paths = PathsFromRoot(root);
sort(paths.begin(), paths.end(), greater<>());
// case 1. only root
R[1] = max(R[1], to_root);
if(paths.empty()) return;
// case 2. root and some leaves
ll cost2 = to_root + paths[0].first;
R[2] = max(R[2], cost2);
for(int i=1; i<paths.size(); i++) R[i+2] = max(R[i+2], cost2 += paths[i].first);
// case 3. only leaves
int idx = find_if(paths.begin(), paths.end(), [&](auto v){ return paths[0].second != v.second; }) - paths.begin();
if(idx != paths.size()){
ll cost3 = to_root + paths[0].first + paths[idx].first;
paths.erase(paths.begin() + idx);
R[2] = max(R[2], cost3);
for(int i=1; i<paths.size(); i++) R[i+2] = max(R[i+2], cost3 += paths[i].first);
}
}
int main(){
ios_base::sync_with_stdio(false); cin.tie(nullptr);
cin >> N; assert(N <= 2000);
for(int i=1; i<N; i++){
int a, b, c, d; cin >> a >> b >> c >> d; Sum += c + d;
G[a].emplace_back(b, c); G[b].emplace_back(a, d);
}
TreeDP(1);
for(int i=1; i<=N; i++) Go(i);
for(int i=2; i<=N; i++) R[i] = max(R[i], R[i-1]);
cin >> Q;
for(int i=1,t; i<=Q; i++) cin >> t, cout << Sum - R[t] << "\n";
}
