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>
#pragma GCC optimize("Ofast")
using namespace std;
const long double pi = 3.141592653589793238;
int MOD = 998244353;
const int N = 1e5 + 10;
const int LOGN = 20;
int n, m, q;
vector<int> graph[N];
vector<int> person[N];
int h[N], par[N][LOGN];
vector<int> dp[N][LOGN];
void add(vector<int>& a, vector<int>& b, int sz) {
  vector<int> temp;
  int i = 0, j = 0;
  for (auto it : b) a.push_back(it);
  sort(a.begin(), a.end());
  while (a.size() > 10) a.pop_back();
}
void dfs(int u, int p, int level) {
  h[u] = level;
  par[u][0] = p;
  for (auto v : graph[u]) {
    if (v == p) continue;
    dfs(v, u, level + 1);
  }
}
int LCA(int u, int v) {
  if (h[v] < h[u]) swap(u, v);
  for (int j = LOGN - 1; j >= 0; j--) {
    if (par[v][j] != -1 && h[par[v][j]] >= h[u]) {
      v = par[v][j];
    }
  }
  if (v == u) return v;
  for (int j = LOGN - 1; j >= 0; j--) {
    if (par[v][j] != par[u][j]) {
      v = par[v][j];
      u = par[u][j];
    }
  }
  return par[u][0];
}
void solve() {
  cin >> n >> m >> q;
  memset(par, -1, sizeof(par));
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    cin >> c;
    dp[c][0].push_back(i);
    if (dp[c][0].size() > 10) dp[c][0].pop_back();
  }
  dfs(1, -1, 1);
  for (int j = 1; j < LOGN; j++) {
    for (int i = 1; i <= n; i++) {
      if (par[i][j - 1] == -1) continue;
      par[i][j] = par[par[i][j - 1]][j - 1];
      add(dp[i][j], dp[par[i][j - 1]][j - 1], 10);
      add(dp[i][j], dp[i][j - 1], 10);
    }
  }
  while (q--) {
    int v, u, a;
    cin >> v >> u >> a;
    int lca = LCA(v, u);
    vector<int> sp;
    for (int j = LOGN - 1; j >= 0; j--) {
      if (par[v][j] != -1 && h[par[v][j]] >= h[lca]) {
        add(sp, dp[v][j], a);
        v = par[v][j];
      }
      if (par[u][j] != -1 && h[par[u][j]] >= h[lca]) {
        add(sp, dp[u][j], a);
        u = par[u][j];
      }
    }
    add(sp, dp[lca][0], a);
    while ((int)sp.size() > a) sp.pop_back();
    int sz = sp.size();
    printf("%d ", sz);
    for (int i = 0; i < sz; i++) printf("%d ", sp[i]);
    printf("\n");
  }
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}