#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 200010;
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pil pair<int, pll>
#define pli pair<ll, int>
#define mp make_pair
int n, q;
ll seg[maxn*4]; //a max seg tree
//we want the seg-tree to go in euler tour order
int qus[maxn];
vector<pil> adj[maxn];
ll ans[maxn]; //store ans for each value
ll dep[maxn]; //distance from root to me
int par[maxn];
ll plink[maxn];
int rt; //some global root variable
ll esum = 0LL; //total edge sum
//barebones dfs
void dfs(int u, int p = -1) {
if (p == -1) {
dep[u] = 0LL;
par[u] = -1;
}
for (pil vp : adj[u]) {
if (vp.first == p) continue;
dep[vp.first] = dep[u] + vp.second.first; //dep is length to bot
par[vp.first] = u;
plink[vp.first] = vp.second.second;
dfs(vp.first, u);
}
}
void dfs1(int u, ll msum = 0LL) {
ans[1] = min(ans[1], esum - msum);
for (pil vp : adj[u]) {
if (vp.first != par[u]) {
dfs1(vp.first, msum - vp.second.second +
vp.second.first);
}
}
}
void go1() {
//finds the answer for 1 by itself
//when I go to a child, I reverse that edge
//start with all going down
ans[1] = esum;
ll csum = 0LL;
for (int i = 1; i <= n; i++) {
if (i != rt) csum += plink[i];
}
dfs1(rt, csum);
}
pli mdepth[maxn]; //want maxdepth
pii cg;
void dfs2(int u, ll msum) {
//msum is the sum of everything in
//consider me to be an lca
mdepth[u] = {dep[u], u};
vector<pli> ops;
for (pil vp : adj[u]) {
if (vp.first == par[u]) continue;
dfs2(vp.first, msum - vp.second.second);
ops.push_back(mdepth[vp.first]);
}
sort(ops.begin(), ops.end());
reverse(ops.begin(), ops.end());
if (ops.size()) mdepth[u] = ops[0];
if (ops.size() < 2) return;
msum += ops[0].first + ops[1].first - dep[u];
if (esum - msum < ans[2]) {
ans[2] = esum - msum;
cg = mp(ops[0].second, ops[1].second);
}
}
pii go2() {
//we consider everything as the lca
//then we sacrifice a certain amount that "goes down"
//start with all of the par-links in (par points up)
ans[2] = esum;
// we lose some of the par-links
ll csum = 0LL;
for (int i = 1; i <= n; i++) {
if (i != rt) csum += plink[i];
}
dfs2(rt, csum);
return cg;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n;
int a, b;
ll c, d;
for (int i = 0; i < n-1; i++) {
cin >> a >> b >> c >> d;
adj[a].emplace_back(b, mp(c, d));
adj[b].emplace_back(a, mp(d, c));
esum += c;
esum += d;
}
cin >> q;
for (int i = 1; i <= q; i++) {
cin >> qus[i];
}
if (n == 2) {
//just bash
for (int i = 1; i <= q; i++) {
if (qus[i] == 2) {
cout << 0 << endl;
}
else {
cout << min(adj[1][0].second.first,
adj[1][0].second.second);
}
}
return 0;
}
//now we want to root at a non-leaf
rt = 1;
for (int i = 2; i <= n; i++) {
if (adj[i].size() != 1) rt = i;
}
dfs(rt);
go1();
pii vp = go2();
//now we are just printing out answer (for now - 1/2)
for (int i = 1; i <= q; i++) {
cout << ans[qus[i]] << '\n';
}
cout.flush();
}
//calculate the answer for one
//use a dp to calculate the answer for two (root at non-leaf)
//greedily add nodes until we get to each val (if none left - do nothing)
