Problem: 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.

<image>

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.

Input

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).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

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


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct node;
typedef node *Node;
const int inf = 1000000;
bool fl = 0;
struct node {
  Node left, right, link, parent;
  int size, prior, sum, number;
  set<int> all;
  pair<int, int> mmin;
  node(int number) : number(number) {
    left = right = link = parent = NULL;
    prior = rand() + (rand() << 16);
    update();
  }
  int key() {
    int ans = 1;
    if (left) ans += left->size;
    return ans;
  }
  void update() {
    size = 1;
    sum = all.size();
    if (sum == 0)
      mmin = make_pair(inf, inf);
    else
      mmin = make_pair(*all.begin(), number);
    if (left) {
      size += left->size;
      sum += left->sum;
      mmin = min(mmin, left->mmin);
      left->parent = this;
    }
    if (right) {
      size += right->size;
      sum += right->sum;
      mmin = min(mmin, right->mmin);
      right->parent = this;
    }
  }
};
void upd_min(Node t) {
  while (t) {
    t->update();
    t = t->parent;
  }
}
void add_ppl(Node &t, int ppl) {
  t->all.insert(ppl);
  t->sum++;
  upd_min(t);
}
void del_ppl(Node &t, int ppl) {
  t->all.erase(ppl);
  t->sum--;
  upd_min(t);
}
Node merge(Node l, Node r) {
  if (!l) return r;
  if (!r) return l;
  if (l->prior > r->prior) {
    l->right = merge(l->right, r);
    l->update();
    return l;
  } else {
    r->left = merge(l, r->left);
    r->update();
    return r;
  }
}
void split(Node t, int key, Node &l, Node &r) {
  if (!t) return void(l = r = NULL);
  if (t->key() <= key) {
    split(t->right, key - t->key(), t->right, r);
    l = t;
  } else {
    split(t->left, key, l, t->left);
    r = t;
  }
  t->update();
  t->parent = NULL;
}
int get_pos(Node t) {
  vector<char> vc;
  Node v = t;
  while (v->parent) {
    if (v->parent->left == v)
      vc.push_back(0);
    else
      vc.push_back(1);
    v = v->parent;
  }
  reverse(vc.begin(), vc.end());
  int pos = 0;
  for (int i = 0; i < vc.size(); ++i) {
    if (vc[i] == 0)
      v = v->left;
    else
      pos += v->key(), v = v->right;
  }
  pos += v->key();
  return pos;
}
Node get_root(Node t) {
  Node ans = NULL;
  while (t) {
    ans = t;
    t = t->parent;
  }
  return ans;
}
Node leftmost(Node t) {
  Node ans = NULL;
  while (t) {
    ans = t;
    t = t->left;
  }
  return ans;
}
Node rightmost(Node t) {
  Node ans = NULL;
  while (t) {
    ans = t;
    t = t->right;
  }
  return ans;
}
void set_link(Node t, Node v) {
  if (!t) return;
  t->link = v;
}
void cut_out(Node t) {
  int pos = get_pos(t);
  Node l, r;
  split(get_root(t), pos, l, r);
  set_link(leftmost(r), t);
}
void cut_out1(Node t) {
  int pos = get_pos(t);
  Node l, r;
  split(get_root(t), pos - 1, l, r);
  set_link(leftmost(r), rightmost(l));
}
void expose(Node v) {
  cut_out(v);
  v = leftmost(get_root(v));
  while (v->link) {
    cut_out(v->link);
    merge(get_root(v->link), get_root(v));
    set_link(v, NULL);
    v = leftmost(get_root(v));
  }
}
void link(Node u, Node v) {
  set_link(u, v);
  expose(u);
}
void cut(Node u, Node v) {
  expose(v);
  set_link(u, NULL);
}
Node lca(Node v, Node u, bool &one_tree) {
  expose(u);
  if (get_root(u) == get_root(v)) {
    one_tree = 1;
    return v;
  }
  expose(v);
  expose(u);
  return leftmost(get_root(v))->link;
}
vector<Node> nodes;
const int maxn = 100000;
vector<pair<int, int> > add;
vector<int> g[maxn + 1];
vector<int> deep;
void dfs(int v, int p = -1) {
  for (int i = 0; i < g[v].size(); ++i) {
    int tv = g[v][i];
    if (tv == p) continue;
    deep[tv] = deep[v] + 1;
    dfs(tv, v);
  }
}
void print_(Node v) {
  if (!v) return;
  print_(v->left);
  cout << "\t" << v->number << " " << (v->link ? v->link->number : -1) << endl;
  print_(v->right);
}
void print(Node v) { print_(get_root(v)); }
void print_all(int n) {
  for (int i = 1; i <= n; ++i) {
    cout << i << ":\n";
    print(nodes[i]);
    cout << endl;
  }
}
void do_the_thing(Node v, int x) {
  Node t = get_root(v);
  int k = min(x, t->sum);
  cout << k << " ";
  vector<pair<int, int> > add_after;
  for (int i = 0; i < k; ++i) {
    pair<int, int> cmin = t->mmin;
    cout << cmin.first << " ";
    del_ppl(nodes[cmin.second], cmin.first);
    add_after.push_back(cmin);
  }
  cout << endl;
  for (int i = 0; i < add_after.size(); ++i) {
    pair<int, int> ct = add_after[i];
    add_ppl(nodes[ct.second], ct.first);
  }
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(0);
  int n, m, q;
  cin >> n >> m >> q;
  nodes.resize(n + 1);
  deep.assign(n + 1, 0);
  for (int i = 1; i <= n; ++i) nodes[i] = new node(i);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    add.push_back(make_pair(a, b));
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  for (int i = 0; i < add.size(); ++i) {
    int a = add[i].first;
    int b = add[i].second;
    if (deep[a] < deep[b]) swap(a, b);
    link(nodes[a], nodes[b]);
  }
  for (int i = 1; i <= m; ++i) {
    int city;
    cin >> city;
    add_ppl(nodes[city], i);
  }
  for (int i = 0; i < q; ++i) {
    int v, u, x;
    cin >> v >> u >> x;
    bool one_tree = 0;
    Node t = lca(nodes[v], nodes[u], one_tree);
    if (one_tree) {
      cut_out1(nodes[v]);
      do_the_thing(nodes[v], x);
    } else {
      int split_key = get_root(nodes[v])->size;
      cut_out1(t);
      merge(get_root(nodes[v]), get_root(t));
      do_the_thing(nodes[v], x);
      Node l, r;
      split(get_root(nodes[v]), split_key, l, r);
    }
  }
  return 0;
}