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 long long maxn = 100000 + 100;
const long long mod = 1e9 + 7;
const long long maxlog = 19;
vector<int> g[maxn], adad[maxn];
vector<int> dp[maxn][maxlog];
int par[maxn][maxlog], h[maxn];
bool visit[maxn];
void dfs(int v, int pp) {
  visit[v] = 1;
  if (pp == -1) {
    h[0] = 0;
    for (int i = 0; i < maxlog; i++) par[0][i] = -1;
    for (int i = 0; i < g[v].size(); i++)
      if (!visit[g[v][i]]) dfs(g[v][i], v);
    return;
  }
  par[v][0] = pp;
  h[v] = h[pp] + 1;
  for (int i = 0; i < adad[v].size(); i++) dp[v][0].push_back(adad[v][i]);
  sort(dp[v][0].begin(), dp[v][0].end());
  while (dp[v][0].size() > 10) dp[v][0].pop_back();
  for (int j = 1; j < maxlog; j++) {
    if (par[v][j - 1] == -1 || par[par[v][j - 1]][j - 1] == -1) {
      par[v][j] = -1;
      continue;
    }
    par[v][j] = par[par[v][j - 1]][j - 1];
    for (int i = 0; i < min((int)dp[v][j - 1].size(), 10); i++)
      dp[v][j].push_back(dp[v][j - 1][i]);
    for (int i = 0; i < min((int)dp[par[v][j - 1]][j - 1].size(), 10); i++)
      dp[v][j].push_back(dp[par[v][j - 1]][j - 1][i]);
    sort(dp[v][j].begin(), dp[v][j].end());
    while (dp[v][j].size() > 10) dp[v][j].pop_back();
  }
  for (int i = 0; i < g[v].size(); i++)
    if (!visit[g[v][i]]) dfs(g[v][i], v);
}
int LCA(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  for (int i = maxlog - 1; i >= 0; i--) {
    if (par[u][i] != -1 && h[par[u][i]] >= h[v]) u = par[u][i];
  }
  if (u == v) return u;
  for (int i = maxlog - 1; i >= 0; i--) {
    if (par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
  }
  return par[u][0];
}
vector<int> query(int u, int v) {
  vector<int> ans;
  if (u == v) return ans;
  int j;
  for (j = maxlog - 1; j >= 0; j--) {
    if (par[u][j] != -1 && h[par[u][j]] >= h[v]) break;
  }
  for (int i = 0; i < min((int)dp[u][j].size(), 10); i++)
    ans.push_back(dp[u][j][i]);
  vector<int> res = query(par[u][j], v);
  for (int i = 0; i < res.size(); i++) ans.push_back(res[i]);
  sort(ans.begin(), ans.end());
  while (ans.size() > 10) ans.pop_back();
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    x--;
    adad[x].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) sort(adad[i].begin(), adad[i].end());
  dfs(0, -1);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    u--;
    v--;
    vector<int> ans, ans2;
    int lca = LCA(u, v);
    if (lca != u) ans = query(u, lca);
    if (lca != v) ans2 = query(v, lca);
    for (int i = 0; i < min((int)adad[lca].size(), 10); i++)
      ans.push_back(adad[lca][i]);
    for (int i = 0; i < ans2.size(); i++) ans.push_back(ans2[i]);
    sort(ans.begin(), ans.end());
    cout << min(a, (int)ans.size()) << ' ';
    for (int i = 0; i < min(a, (int)ans.size()); i++) cout << ans[i] << ' ';
    cout << endl;
  }
  return 0;
}