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;
vector<int> a[200000], b[200000], cha[200000][40], res;
int n, m, q, cha2[200000][40], d[200000];
void DFS(int u, int p) {
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (p == v) continue;
    int tmp = b[v].size();
    tmp = min(tmp, 10);
    cha2[v][0] = u;
    d[v] = d[u] + 1;
    for (int i = 0; i < tmp; i++) cha[v][0].push_back(b[v][i]);
    DFS(v, u);
  }
}
void solve(int u, int v, int k) {
  if (d[u] > d[v]) swap(u, v);
  int kc = d[v] - d[u];
  for (int i = 0; i <= 30; i++)
    if ((kc >> i) & 1) {
      for (int j = 0; j < cha[v][i].size(); j++) res.push_back(cha[v][i][j]);
      sort(res.begin(), res.end());
      res.resize(min(k, (int)res.size()));
      v = cha2[v][i];
    }
  for (int i = 30; i >= 1; i--)
    if (cha2[v][i] != cha2[u][i]) {
      for (int j = 0; j < cha[v][i].size(); j++) res.push_back(cha[v][i][j]);
      for (int j = 0; j < cha[u][i].size(); j++) res.push_back(cha[u][i][j]);
      sort(res.begin(), res.end());
      res.resize(min(k, (int)res.size()));
      v = cha2[v][i], u = cha2[u][i];
    }
  while (u != v) {
    for (int j = 0; j < cha[v][0].size(); j++) res.push_back(cha[v][0][j]);
    for (int j = 0; j < cha[u][0].size(); j++) res.push_back(cha[u][0][j]);
    sort(res.begin(), res.end());
    res.resize(min(k, (int)res.size()));
    v = cha2[v][0], u = cha2[u][0];
  }
  for (int j = 0; j < cha[v][0].size(); j++) res.push_back(cha[v][0][j]);
  sort(res.begin(), res.end());
  res.resize(min(k, (int)res.size()));
  v = cha2[v][0], u = cha2[u][0];
}
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    a[u].push_back(v);
    a[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int tmp;
    cin >> tmp;
    b[tmp].push_back(i);
  }
  for (int i = 0; i < min((int)b[1].size(), 10); i++)
    cha[1][0].push_back(b[1][i]);
  d[1] = 0;
  DFS(1, 0);
  for (int j = 1; j <= 30; j++)
    for (int i = 1; i <= n; i++) {
      int tmp = cha2[i][j - 1];
      int tmp2 = cha[i][j - 1].size() + cha[tmp][j - 1].size();
      tmp2 = min(tmp2, 10);
      int x = 0, y = 0;
      int len1 = cha[i][j - 1].size(), len2 = cha[tmp][j - 1].size();
      for (int o = 0; o < tmp2; o++) {
        if (x < len1 and (y == len2 or cha[i][j - 1][x] < cha[tmp][j - 1][y]))
          cha[i][j].push_back(cha[i][j - 1][x]), x++;
        else
          cha[i][j].push_back(cha[tmp][j - 1][y]), y++;
      }
      cha2[i][j] = cha2[cha2[i][j - 1]][j - 1];
    }
  for (int i = 1; i <= q; i++) {
    int u, v, k;
    cin >> u >> v >> k;
    res.clear();
    solve(u, v, k);
    cout << res.size() << " ";
    for (int i = 0; i < res.size(); i++) cout << res[i] << " ";
    cout << "\n";
  }
}