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;
int N, M, Q;
int depth[100000 + 1];
int up[100000 + 1][20 + 1];
vector<int> people[100000 + 1][20 + 1];
vector<int> adj[100000 + 1];
vector<int> merge(vector<int> a, vector<int> b, int sz = 10) {
  vector<int> ret;
  int i = 0, j = 0;
  while (ret.size() <= sz && (i < a.size() || j < b.size())) {
    if (i == a.size() || (j != b.size() && b[j] < a[i]))
      ret.push_back(b[j++]);
    else
      ret.push_back(a[i++]);
  }
  return ret;
}
void dfs(int node, int par) {
  depth[node] = depth[up[node][0] = par] + 1;
  for (int i = 1; i <= 20; i++) {
    people[node][i] =
        merge(people[node][i - 1], people[up[node][i - 1]][i - 1]);
    up[node][i] = up[up[node][i - 1]][i - 1];
  }
  for (int i : adj[node])
    if (i != par) dfs(i, node);
}
pair<int, vector<int>> jmp(int x, int k) {
  vector<int> ret;
  for (int i = 0; i <= 20; i++)
    if ((k >> i) & 1) ret = merge(ret, people[x][i]), x = up[x][i];
  return {x ? x : -1, ret};
}
vector<int> solve(int a, int b) {
  if (depth[a] < depth[b]) swap(a, b);
  vector<int> ret;
  pair<int, vector<int>> ans = jmp(a, depth[a] - depth[b]);
  a = ans.first, ret = ans.second;
  if (a == b) return merge(ret, people[a][0]);
  for (int i = 20; i >= 0; i--) {
    int newA = up[a][i], newB = up[b][i];
    if (newA != newB) {
      ret = merge(ret, merge(people[a][i], people[b][i]));
      a = newA, b = newB;
    }
  }
  return merge(ret,
               merge(merge(people[a][0], people[b][0]), people[up[a][0]][0]));
}
int main() {
  cin.tie(0)->sync_with_stdio(false);
  cin >> N >> M >> Q;
  for (int i = 2; i <= N; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= M; i++) {
    int x;
    cin >> x;
    people[x][0].push_back(i);
  }
  for (int i = 1; i <= N; i++) sort(people[i][0].begin(), people[i][0].end());
  dfs(1, 0);
  for (int i = 1; i <= Q; i++) {
    int u, v, a;
    cin >> u >> v >> a;
    vector<int> ans = solve(u, v);
    int k = min((int)ans.size(), a);
    cout << k << " ";
    for (int i = 0; i < k; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}