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;
void solve();
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
const int N = 100001, M = 18, K = 10;
int n, m, q;
vector<int> g[N];
bool used[N];
int tin[N], tout[N], tctr;
int p[N][M];
vector<int> res[N][M];
void union_(const vector<int>& a, const vector<int>& b, vector<int>& res,
            int k = K) {
  int n1 = a.size(), n2 = b.size();
  res.resize(n1 + n2);
  auto it = set_union(a.begin(), a.end(), b.begin(), b.end(), res.begin());
  res.resize(min(k, (int)(it - res.begin())));
}
void dfs(int u, int par = -1) {
  used[u] = true;
  tin[u] = tctr++;
  p[u][0] = (par == -1 ? u : par);
  for (int i = 1; i < M; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
    union_(res[u][i - 1], res[p[u][i - 1]][i - 1], res[u][i]);
  }
  for (int v : g[u])
    if (v != par) dfs(v, u);
  tout[u] = tctr++;
}
bool is_par(int u, int v) { return tin[u] <= tin[v] && tout[u] >= tout[v]; }
vector<int> query(int u, int v, int k) {
  vector<int> ret;
  for (int i = M - 1; i >= 0; --i) {
    if (!is_par(p[u][i], v)) {
      vector<int> tmp;
      union_(ret, res[u][i], tmp, k);
      ret.swap(tmp);
      u = p[u][i];
    }
    if (!is_par(p[v][i], u)) {
      vector<int> tmp;
      union_(ret, res[v][i], tmp, k);
      ret.swap(tmp);
      v = p[v][i];
    }
  }
  int lca = is_par(u, v) ? u : p[u][0];
  vector<int> tmp;
  union_(ret, res[lca][0], tmp, k);
  union_(res[u][0], tmp, ret, k);
  union_(res[v][0], ret, tmp, k);
  return tmp;
}
void solve() {
  while (cin >> n >> m >> q) {
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    for (int i = 0; i < m; i++) {
      int x;
      cin >> x;
      res[x][0].push_back(i + 1);
    }
    for (int i = 1; i <= n; i++) {
      sort(res[i][0].begin(), res[i][0].end());
      if (res[i][0].size() > K) res[i][0].resize(K);
    }
    dfs(1);
    for (int i = 0; i < q; i++) {
      int u, v, k;
      cin >> u >> v >> k;
      vector<int> res = query(u, v, k);
      cout << res.size() << " ";
      for (int x : res) cout << x << " ";
      cout << endl;
    }
  }
}