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, u, v;
pair<int, vector<int> > P[100011][25];
vector<int> residents[100011];
vector<int> adj[100001];
int parent[100001];
int L[100001];
int tin[100001], tout[100001], ti = 0;
int checker;
vector<int> mergevec(vector<int>& A, vector<int>& B, int sz) {
  vector<int> sol;
  int ap = 0, bp = 0;
  while (sz != 0 && ap != A.size() && bp != B.size()) {
    if (A[ap] < B[bp]) {
      if (sol.size() == 0 || sol.back() != A[ap]) sol.push_back(A[ap]), sz--;
      ap++;
    } else {
      if (sol.size() == 0 || sol.back() != B[bp]) sol.push_back(B[bp]), sz--;
      bp++;
    }
  }
  while (sz != 0 && ap != A.size()) {
    if (sol.size() == 0 || sol.back() != A[ap]) sol.push_back(A[ap]), sz--;
    ap++;
  }
  while (sz != 0 && bp != B.size()) {
    if (sol.size() == 0 || sol.back() != B[bp]) sol.push_back(B[bp]), sz--;
    bp++;
  }
  return sol;
}
bool upper(int a, int b) { return (tin[a] <= tin[b] && tout[a] >= tout[b]); }
void process() {
  int i, j;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= 20; j++) {
      P[i][j].first = -1;
    }
  }
  for (i = 1; i <= n; i++) {
    P[i][0].first = parent[i];
    P[i][0].second = residents[i];
  }
  for (j = 1; 1 << j < n; j++) {
    for (i = 1; i <= n; i++) {
      if (P[i][j - 1].first != -1) {
        P[i][j].first = P[P[i][j - 1].first][j - 1].first;
        P[i][j].second = mergevec(P[i][j - 1].second,
                                  P[P[i][j - 1].first][j - 1].second, 10);
      }
    }
  }
}
int lca(int p, int q) {
  if (upper(p, q)) return p;
  if (upper(q, p)) return q;
  for (int i = 17; i >= 0; i--) {
    if (P[p][i].first != -1 && (!upper(P[p][i].first, q))) {
      p = P[p][i].first;
    }
  }
  assert(p != q);
  return P[p][0].first;
}
vector<int> path(int u, int levelpar, int sz) {
  vector<int> sol = residents[u];
  for (int i = checker; i >= 0; i--) {
    if (P[u][i].first != -1) {
      if (L[P[u][i].first] >= levelpar) {
        sol = mergevec(sol, P[u][i].second, sz);
        u = P[u][i].first;
      }
    }
  }
  return sol;
}
void dfs(int start, int p) {
  parent[start] = p;
  tin[start] = ti++;
  if (p == -1)
    L[start] = 1;
  else
    L[start] = L[p] + 1;
  for (int i = 0; i < adj[start].size(); i++) {
    if (adj[start][i] != p) {
      dfs(adj[start][i], start);
    }
  }
  tout[start] = ti++;
}
int main() {
  int a;
  scanf("%d%d%d", &n, &m, &q);
  checker = min((int)(log2(n) + 3), (int)20);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &u);
    if (residents[u].size() < 10) residents[u].push_back(i + 1);
  }
  for (int i = 1; i <= 100000; i++)
    sort(residents[i].begin(), residents[i].end());
  dfs(1, -1);
  process();
  for (int i = 0; i < q; i++) {
    scanf("%d%d%d", &u, &v, &a);
    int lcaa = lca(u, v);
    vector<int> v1 = path(u, L[lcaa], a), v2 = path(v, L[lcaa], a);
    vector<int> sol = mergevec(v1, v2, a);
    sol = mergevec(sol, residents[lcaa], a);
    printf("%d ", (int)sol.size());
    for (int j = 0; j < sol.size(); j++) {
      printf("%d ", sol[j]);
    }
    printf("\n");
  }
  return 0;
}