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;
const int n_ = 1e5 + 10, LOG = 16;
int n, m, q;
vector<int> tree[n_];
int lvl[n_], parentLog[n_][LOG + 1];
vector<int> residents[n_], resLog[n_][LOG + 1];
void dfs(int node = 0, int p = -1) {
  parentLog[node][0] = p;
  for (int i = 0; i < (int)tree[node].size(); i++) {
    int child = tree[node][i];
    if (child != p) {
      lvl[child] = lvl[node] + 1;
      dfs(child, node);
    }
  }
}
int LCA(int u, int v) {
  if (lvl[u] < lvl[v]) swap(u, v);
  for (int log = LOG; log >= 0; log--)
    if (lvl[u] - (1 << log) >= lvl[v]) u = parentLog[u][log];
  if (u == v) return u;
  for (int log = LOG; log >= 0; log--)
    if (parentLog[u][log] != -1 && parentLog[u][log] != parentLog[v][log])
      u = parentLog[u][log], v = parentLog[v][log];
  return parentLog[u][0];
}
vector<int> query(int u, int lca) {
  vector<int> res;
  for (int log = LOG; log >= 0; log--)
    if (lvl[u] - (1 << log) >= lvl[lca]) {
      for (int i = 0; i < (int)resLog[u][log].size(); i++)
        res.push_back(resLog[u][log][i]);
      u = parentLog[u][log];
    }
  for (int i = 0; i < (int)residents[u].size(); i++)
    res.push_back(residents[u][i]);
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  memset(parentLog, -1, sizeof parentLog);
  dfs();
  for (int log = 1; log <= LOG; log++)
    for (int i = 0; i < n; i++)
      if (parentLog[i][log - 1] != -1)
        parentLog[i][log] = parentLog[parentLog[i][log - 1]][log - 1];
  for (int i = 0; i < m; i++) {
    int c;
    cin >> c;
    c--;
    if ((int)residents[c].size() < 10) residents[c].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) sort(residents[i].begin(), residents[i].end());
  resLog[0][0] = residents[0];
  for (int i = 1; i < n; i++) {
    int x, y;
    for (x = 0, y = 0; x < (int)residents[parentLog[i][0]].size() &&
                       y < (int)residents[i].size() && x + y < 10;)
      if (residents[parentLog[i][0]][x] < residents[i][y])
        resLog[i][0].push_back(residents[parentLog[i][0]][x++]);
      else
        resLog[i][0].push_back(residents[i][y++]);
    while (x + y < 10 && x < (int)residents[parentLog[i][0]].size())
      resLog[i][0].push_back(residents[parentLog[i][0]][x++]);
    while (x + y < 10 && y < (int)residents[i].size())
      resLog[i][0].push_back(residents[i][y++]);
  }
  for (int log = 1; log <= LOG; log++) {
    for (int i = 0; i < n; i++) {
      if (parentLog[i][log - 1] == -1) continue;
      int x, y;
      set<int> taken;
      for (x = 0, y = 0;
           x < (int)resLog[i][log - 1].size() &&
           y < (int)resLog[parentLog[i][log - 1]][log - 1].size() &&
           (int)taken.size() < 10;)
        if (resLog[i][log - 1][x] < resLog[parentLog[i][log - 1]][log - 1][y])
          taken.insert(resLog[i][log - 1][x]),
              resLog[i][log].push_back(resLog[i][log - 1][x++]);
        else
          taken.insert(resLog[parentLog[i][log - 1]][log - 1][y]),
              resLog[i][log].push_back(
                  resLog[parentLog[i][log - 1]][log - 1][y++]);
      while ((int)taken.size() < 10 && x < (int)resLog[i][log - 1].size())
        taken.insert(resLog[i][log - 1][x]),
            resLog[i][log].push_back(resLog[i][log - 1][x++]);
      while ((int)taken.size() < 10 &&
             y < (int)resLog[parentLog[i][log - 1]][log - 1].size())
        taken.insert(resLog[parentLog[i][log - 1]][log - 1][y]),
            resLog[i][log].push_back(
                resLog[parentLog[i][log - 1]][log - 1][y++]);
    }
  }
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    u--, v--;
    int lca = LCA(u, v);
    vector<int> v1 = query(u, lca), v2 = query(v, lca);
    sort(v1.begin(), v1.end());
    sort(v2.begin(), v2.end());
    int x, y;
    set<int> st;
    for (x = 0, y = 0;
         (int)st.size() < a && x < (int)v1.size() && y < (int)v2.size();)
      if (v1[x] < v2[y])
        st.insert(v1[x++]);
      else
        st.insert(v2[y++]);
    while ((int)st.size() < a && x < (int)v1.size()) st.insert(v1[x++]);
    while ((int)st.size() < a && y < (int)v2.size()) st.insert(v2[y++]);
    cout << (int)st.size();
    for (set<int>::iterator it = st.begin(); it != st.end(); it++)
      cout << " " << *it;
    cout << "\n";
  }
}