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, c[200010];
int F[100010][22], dep[100010];
vector<int> g[100010], s[100010][22];
inline void Combine(vector<int> &a, vector<int> &b, vector<int> &c) {
  int head1 = 0, head2 = 0;
  a.clear();
  while (head1 < b.size() && head2 < c.size()) {
    if (b[head1] < c[head2]) {
      a.push_back(b[head1]);
      head1++;
    } else if (c[head2] < b[head1]) {
      a.push_back(c[head2]);
      head2++;
    } else {
      head1++;
      head2++;
    }
    if (a.size() == 10) return;
  }
  while (head1 < b.size()) {
    a.push_back(b[head1]);
    head1++;
  }
  while (head2 < c.size()) {
    a.push_back(c[head2]);
    head2++;
  }
  if (a.size() > 10) a.resize(10);
}
inline void Combine(vector<int> &a, vector<int> &b) {
  vector<int> tmp = a;
  Combine(a, tmp, b);
}
void DFS(int u) {
  for (int i = 1; i <= 18 - 1; i++) {
    F[u][i] = F[F[u][i - 1]][i - 1];
    Combine(s[u][i], s[u][i - 1], s[F[u][i - 1]][i - 1]);
  }
  for (int i = 0; i < g[u].size(); i++)
    if (!dep[g[u][i]]) {
      dep[g[u][i]] = dep[u] + 1;
      F[g[u][i]][0] = u;
      DFS(g[u][i]);
    }
}
inline void Query(int u, int v, int w) {
  vector<int> total;
  total.clear();
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 18 - 1; i >= 0; i--)
    if (dep[F[u][i]] >= dep[v]) {
      Combine(total, s[u][i]);
      u = F[u][i];
    }
  for (int i = 18 - 1; i >= 0; i--)
    if (F[u][i] != F[v][i]) {
      Combine(total, s[u][i]);
      Combine(total, s[v][i]);
      u = F[u][i];
      v = F[v][i];
    }
  if (u != v) {
    if (dep[u] == dep[v]) Combine(total, s[u][0]);
    Combine(total, s[v][0]);
    u = F[u][0];
    v = F[v][0];
  }
  if (u) Combine(total, s[u][0]);
  if (total.size() > w) total.resize(w);
  printf("%d", total.size());
  for (int i = 0; i < total.size(); i++) printf(" %d", total[i]);
  printf("\n");
}
inline bool cmp(int x, int y) { return x < y; }
inline void solve() {
  int u, v, w;
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &c[i]);
    s[c[i]][0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(s[i][0].begin(), s[i][0].end(), cmp);
    if (s[i][0].size() > 10) s[i][0].resize(10);
  }
  dep[1] = 1;
  F[1][0] = 0;
  DFS(1);
  for (int i = 1; i <= Q; i++) {
    scanf("%d%d%d", &u, &v, &w);
    Query(u, v, w);
  }
}
int main() {
  solve();
  return 0;
}