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, par[20][100005], h[100005];
vector<int> city[100005], a[100005], people[20][100005];
vector<int> Merge(vector<int> a, vector<int> b) {
  int n = a.size(), m = b.size(), i = 0, j = 0;
  while (n <= 10) {
    a.push_back(INT_MAX);
    n++;
  }
  while (m <= 10) {
    b.push_back(INT_MAX);
    m++;
  }
  vector<int> c;
  while (i < n) {
    while (j < m && a[i] > b[j]) {
      c.push_back(b[j]);
      j++;
      if (i + j == 10) break;
    }
    if (i + j == 10) break;
    c.push_back(a[i]);
    i++;
    if (i + j == 10) break;
  }
  return c;
}
void dfs(int u) {
  for (auto v : a[u]) {
    if (v == par[0][u]) continue;
    h[v] = h[u] + 1;
    par[0][v] = u;
    for (auto x : city[v]) people[0][v].push_back(x);
    for (int i = 1; i <= 17; i++) {
      par[i][v] = par[i - 1][par[i - 1][v]];
      people[i][v] = Merge(people[i - 1][v], people[i - 1][par[i - 1][v]]);
    }
    dfs(v);
  }
}
void Out(vector<int> res, int cnt) {
  vector<int> ans;
  for (auto x : res)
    if (x != INT_MAX && ans.size() < cnt)
      ans.push_back(x);
    else
      break;
  cout << ans.size() << " ";
  for (auto x : ans) cout << x << " ";
  cout << '\n';
}
void kruskal(int u, int v, int x) {
  vector<int> res;
  if (h[u] < h[v]) swap(u, v);
  int len = h[u] - h[v];
  for (int i = 17; i >= 0; i--)
    if ((len >> i) & 1) {
      res = Merge(people[i][u], res);
      u = par[i][u];
    }
  if (u == v) {
    res = Merge(people[0][u], res);
    Out(res, x);
    return;
  }
  for (int i = 17; i >= 0; i--)
    if (par[i][u] != par[i][v]) {
      res = Merge(res, people[i][u]);
      res = Merge(res, people[i][v]);
      u = par[i][u];
      v = par[i][v];
    }
  res = Merge(res, people[0][u]);
  res = Merge(res, people[1][v]);
  Out(res, x);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  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 a;
    cin >> a;
    city[a].push_back(i);
  }
  for (int i = 1; i <= n; i++) sort(city[i].begin(), city[i].end());
  for (auto x : city[1]) people[0][1].push_back(x);
  dfs(1);
  while (q--) {
    int u, v, x;
    cin >> u >> v >> x;
    kruskal(u, v, x);
  }
  return 0;
}