#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define F first
#define S second
#define all(x) x.begin(),x.end()
#define pii pair<int,int>
#define pb push_back
#define sz(x) (int)(x.size())
#define chmin(x,y) x=x<y?x:y
#define chmax(x,y) x=x>y?x:y
const int mxn=2e5+5;
inline char gc()
{
const static int SZ = 1 << 16;
static char buf[SZ], *p1, *p2;
if (p1 == p2 && (p2 = buf + fread(p1 = buf, 1, SZ, stdin), p1 == p2))
return -1;
return *p1++;
}
inline void rd() {}
template <typename T, typename... U>
inline void rd(T &x, U &...y)
{
x = 0;
bool f = 0;
char c = gc();
while (isdigit(c))
x = (x << 1) + (x << 3) + (c ^ 48), c = gc();
f && (x = -x), rd(y...);
}
template <typename T>
inline void prt(T x)
{
if (x > 9)
prt(x / 10);
putchar((x % 10) ^ 48);
}
vector<tuple<int,int,int>>adj[mxn];
ll dp[mxn][2], dep[mxn];
pair<ll,int> dp2[mxn];
int p1;
inline void dfs(int v,int pa){
for(auto [u,d,d2]:adj[v]){
if(u==pa)continue;
dp[u][0]+=dp[v][0]+d;
dep[u]=dep[v]+d;
dfs(u,v);
dp[v][1]+=d2+dp[u][1];
}
}
ll all,ans[mxn];
inline void calc(int v,int pa){
ll mx=0;
dp2[v].F=dep[v];
dp2[v].S=v;
int t1=0;
for(auto [u,d,d2]:adj[v]){
if(u==pa)continue;
dp[u][1]+=dp[v][1]-dp[u][1]-d2;
calc(u,v);
if(dp2[v].F+dp2[u].F-2*dep[v]>mx){
t1=dp2[v].S;
mx=dp2[v].F+dp2[u].F-2*dep[v];
}
chmax(dp2[v],dp2[u]);
}
chmin(ans[1],all-dp[v][0]-dp[v][1]);
if(ans[2]>all-dp[v][0]-dp[v][1]-mx){
p1=t1;
ans[2]=all-dp[v][0]-dp[v][1]-mx;
}
}
ll len[mxn];
int idx;
inline ll dfs2(int v,int pa){
vector<ll>temp;
for(auto [u,d,d2]:adj[v]){
if(u==pa)continue;
temp.pb(dfs2(u,v)+d);
}
if(sz(temp)==0)return 0;
auto it=max_element(temp.begin(),temp.end());
for(auto it2=temp.begin();it2!=temp.end();it2++){
if(it2==it)continue;
len[idx++]=*it2;
}
return *it;
}
signed main(){
int n,q;
rd(n);
ans[1] = ans[2]=1e18;
for(int i=1;i<n;i++){
int a,b,c,d;
rd(a),rd(b),rd(c),rd(d);
adj[a].pb({b,c,d});
adj[b].pb({a,d,c});
all+=c+d;
}
dfs(1,1);
calc(1,1);
ll v=dfs2(p1,p1);
len[idx++]=v;
sort(len,len+idx,greater<ll>());
for(int i=1;i<idx;i++){
ans[2+i]=ans[1+i]-len[i];
}
rd(q);
while(q--){
int x;
rd(x);
prt(ans[x]);
putchar('\n');
}
}
/*
input:
*/
