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 + 100, LOG = 18, A = 10, inF = 2e9;
int n, m, q, u, v, a, LCA, h[N], upper[N][LOG];
vector<int> nei[N], in[N], vec[N][LOG];
void dfs(int v) {
  vector<int> help;
  for (int i = 1; i < LOG; i++) {
    upper[v][i] = upper[upper[v][i - 1]][i - 1];
    help.clear();
    for (int num : vec[v][i - 1]) help.push_back(num);
    for (int num : vec[upper[v][i - 1]][i - 1]) help.push_back(num);
    sort(help.begin(), help.end());
    help.push_back(inF);
    for (int j = 0; j < help.size() - 1 && vec[v][i].size() <= A; j++)
      if (help[j] != help[j + 1]) vec[v][i].push_back(help[j]);
  }
  for (int u : nei[v])
    if (u != upper[v][0]) {
      h[u] = h[v] + 1;
      upper[u][0] = v;
      help.clear();
      for (int num : in[v]) help.push_back(num);
      for (int num : in[u]) help.push_back(num);
      sort(help.begin(), help.end());
      help.push_back(inF);
      for (int j = 0; j < help.size() - 1 && vec[u][0].size() <= A; j++)
        if (help[j] != help[j + 1]) vec[u][0].push_back(help[j]);
      dfs(u);
    }
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  for (int i = LOG - 1; i >= 0; i--)
    if (h[u] - (1LL << i) >= h[v]) u = upper[u][i];
  if (u == v) return u;
  for (int i = LOG - 1; i >= 0; i--)
    if (upper[u][i] != upper[v][i]) {
      u = upper[u][i];
      v = upper[v][i];
    }
  return upper[u][0];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    nei[--u].push_back(--v);
    nei[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    cin >> u;
    if (in[--u].size() < A) in[u].push_back(i + 1);
  }
  dfs(0);
  set<int> ST;
  while (q--) {
    ST.clear();
    cin >> u >> v >> a;
    LCA = lca(--u, --v);
    for (int num : in[v]) {
      ST.insert(num);
      if (ST.size() > a) ST.erase(--ST.end());
    }
    for (int num : in[u]) {
      ST.insert(num);
      if (ST.size() > a) ST.erase(--ST.end());
    }
    for (int i = LOG - 1; i >= 0; i--)
      if (h[u] - (1LL << i) >= h[LCA]) {
        for (int num : vec[u][i]) {
          ST.insert(num);
          if (ST.size() > a) ST.erase(--ST.end());
        }
        u = upper[u][i];
      }
    for (int i = LOG - 1; i >= 0; i--)
      if (h[v] - (1LL << i) >= h[LCA]) {
        for (int num : vec[v][i]) {
          ST.insert(num);
          if (ST.size() > a) ST.erase(--ST.end());
        }
        v = upper[v][i];
      }
    a = min(a, (int)ST.size());
    cout << a << ' ';
    while (ST.size() > a) ST.erase(--ST.end());
    for (int num : ST) cout << num << ' ';
    cout << '\n';
  }
  return 0;
}