Recently Duff has been a soldier in the army. Malek is her commander.
Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.
There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.
<center>
</center> Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.
To answer a query:
Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order.
If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).
Duff is very busy at the moment, so she asked you to help her and answer the queries.
The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).
The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).
Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).
Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).
For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.
5 4 5 1 3 1 2 1 4 4 5 2 1 4 3 4 5 6 1 5 2 5 5 10 2 3 3 5 3 1
1 3 2 2 3 0 3 1 2 4 1 2
Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):
【题意】给N<=100000的一颗树，M<=100000个人，Q<=100000的询问，每个人都在某个节点，一个节点可以有不定数的人，询问(u,v)上前a个编号的人是哪些？
【解题方法】倍增LCA,have[i][u]，u向上爬2^i方步有多少个人，只需要维护前10个就好了，这里不能用stl的vector，会t哭，所以手写vector，但是我这个效率也很低。测了一份别人的代码，只需要500ms，我这份代码跑了2000多ms。不过毕竟也能过，按照lca往上爬，维护答案就可以了。
【AC 代码】
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n,m,q;
struct Myvec{
int a[10],sz;
Myvec(){sz=0;}
int size(){return sz;}
void push_back(int &x){if(sz<10) a[sz++]=x;}
int &operator[](const int &i){return a[i];}
};
Myvec operator+(Myvec a,Myvec b){
Myvec res;
int i,j;
for(i=j=0; i<a.size()&&j<b.size()&&res.size()<10;){
if(a[i]<b[j]) res.push_back(a[i++]);
else res.push_back(b[j++]);
}
while(res.size()<10&&i<a.size()) res.push_back(a[i++]);
while(res.size()<10&&j<b.size()) res.push_back(b[j++]);
return res;
}
vector<int>G[maxn];
Myvec c[maxn],have[17][maxn];
int dep[maxn],p[17][maxn];
void dfs(int u,int f)
{
p[0][u]=f;
have[0][u]=c[f];
for(int i=1; (1<<i)<=dep[u]; i++){
p[i][u]=p[i-1][p[i-1][u]];
have[i][u]=have[i-1][u]+have[i-1][p[i-1][u]];
}
for(int i=0; i<G[u].size(); i++){
int v=G[u][i];
if(v==f) continue ;
dep[v] = dep[u]+1;
dfs(v,u);
}
}
int lca(int u,int v)
{
if(dep[u]>dep[v]) swap(u,v);
for(int i=0; i<17; i++)
if(dep[v]-dep[u]>>i&1) v=p[i][v];
if(u==v) return u;
for(int i=16; i>=0; i--){
if(p[i][u]!=p[i][v]){
u=p[i][u];
v=p[i][v];
}
}
return p[0][u];
}
Myvec gohead(int u,int _lca)
{
Myvec ret=c[u];
for(int i=0; i<17; i++){
if(dep[u]-dep[_lca]-1>0 && dep[u]-dep[_lca]-1>>i&1){
ret=ret+have[i][u];u=p[i][u];
}
}
return ret;
}
int main()
{
int u,v,a,x;
cin>>n>>m>>q;
for(int i=1; i<n; i++){
cin>>u>>v;
G[u].push_back(v);
G[v].push_back(u);
}
for(int i=1; i<=m; i++)
{
scanf("%d",&x);
if(c[x].size()<10) c[x].push_back(i);
}
dfs(1,0);
while(q--)
{
cin>>u>>v>>a;
int _lca=lca(u,v);
Myvec ans;
if(u!=_lca&&v!=_lca) ans=c[_lca];
ans=ans+gohead(u,_lca);
if(v!=u) ans=ans+gohead(v,_lca);
int k=min(ans.size(),a);
printf("%d ",k);
for(int i=0; i<k; i++){
printf("%d ",ans[i]);
}
printf("\n");
}
return 0;
}
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