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 + 5;
int n, m, q;
vector<int> adj[N];
vector<int> people[N];
int tt;
int st[N], en[N];
int lca[19][N];
vector<int> p[19][N];
vector<int> merge(vector<int> &a, vector<int> &b) {
  vector<int> ret = a;
  ret.insert(ret.end(), b.begin(), b.end());
  sort(ret.begin(), ret.end());
  ret.erase(unique(ret.begin(), ret.end()), ret.end());
  if (ret.size() > 10) ret.resize(10);
  return ret;
}
void dfs(int node, int par) {
  st[node] = ++tt;
  lca[0][node] = par;
  for (int i = 1; i < 19; i++) {
    lca[i][node] = lca[i - 1][lca[i - 1][node]];
  }
  vector<int> v = merge(people[node], people[par]);
  p[0][node] = v;
  for (int i = 1; i < 19; i++) {
    p[i][node] = merge(p[i - 1][node], p[i - 1][lca[i - 1][node]]);
  }
  for (auto to : adj[node]) {
    if (to != par) dfs(to, node);
  }
  en[node] = ++tt;
}
bool check(int a, int b) { return st[a] < st[b] && en[a] > en[b]; }
void output(int l, int r, int a) {
  if (l == r) {
    cout << min(a, (int)people[l].size()) << " ";
    for (int i = 0; i < min(a, (int)people[l].size()); i++) {
      cout << people[l][i] << " ";
    }
    cout << "\n";
    return;
  }
  vector<int> v;
  for (int i = 18; ~i; i--) {
    if (!check(lca[i][l], r)) {
      v = merge(v, p[i][l]);
      l = lca[i][l];
    }
  }
  for (int i = 18; ~i; i--) {
    if (!check(lca[i][r], l)) {
      v = merge(v, p[i][r]);
      r = lca[i][r];
    }
  }
  if (l != r) {
    v = merge(v, p[0][l]);
    v = merge(v, p[0][r]);
  }
  cout << min((int)v.size(), a) << " ";
  for (int i = 0; i < min((int)v.size(), a); i++) cout << v[i] << " ";
  cout << "\n";
}
void solve() {
  cin >> n >> m >> q;
  for (int i = 1; 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 a;
    cin >> a;
    people[a].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(people[i].begin(), people[i].end());
    if (people[i].size() > 10) people[i].resize(10);
  }
  dfs(1, 1);
  while (q--) {
    int l, r, a;
    cin >> l >> r >> a;
    output(l, r, a);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  solve();
  return 0;
}