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 mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const double eps = 1e-9;
int timer = 0;
int n, m, q, k = 10;
vector<vector<int>> g;
int up[25][100005];
vector<int> a[100005], val[25][100005];
int in[100005], out[100005];
vector<int> operator+(const vector<int> &x, const vector<int> &y) {
  vector<int> res(x.size() + y.size());
  merge(x.begin(), x.end(), y.begin(), y.end(), res.begin());
  if (res.size() > k) res.resize(k);
  return res;
}
void dfs(int s, int p) {
  up[0][s] = p;
  if (s != p) val[0][s] = a[p];
  for (int i = 1; i <= 20; ++i) {
    up[i][s] = up[i - 1][up[i - 1][s]];
    val[i][s] = val[i - 1][s] + val[i - 1][up[i - 1][s]];
  }
  in[s] = ++timer;
  for (auto v : g[s]) {
    if (v != p) dfs(v, s);
  }
  out[s] = ++timer;
}
bool anc(int u, int v) { return (in[u] <= in[v] && out[u] >= out[v]); }
int lca(int u, int v) {
  if (anc(u, v)) return u;
  if (anc(v, u)) return v;
  for (int i = 20; i >= 0; --i) {
    if (!anc(up[i][u], v)) u = up[i][u];
  }
  return up[0][u];
}
vector<int> jump(int u, int v) {
  if (u == v) return {};
  vector<int> res = a[u];
  for (int i = 20; i >= 0; --i) {
    if (!anc(up[i][u], v)) res = res + val[i][u], u = up[i][u];
  }
  return res;
}
void solve() {
  cin >> n >> m >> q;
  g.resize(n + 1);
  for (int i = 1, a, b; i <= n - 1; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1, x; i <= m; ++i) {
    cin >> x;
    if (a[x].size() < k) a[x].push_back(i);
  }
  dfs(1, 1);
  while (q--) {
    int u, v;
    cin >> u >> v >> k;
    int par = lca(u, v);
    vector<int> res = jump(u, par) + jump(v, par) + a[par];
    cout << res.size();
    for (auto x : res) cout << " " << x;
    cout << "\n";
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  for (int tc = 1; tc <= test; ++tc) solve();
  return 0;
}