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;
vector<vector<int>> adj(100000), nums(100000);
int N, T[100000], L[100000];
struct {
  int t;
  vector<int> prn;
} P[100000][17];
vector<int> mergeVec(vector<int> const& a, vector<int> const& b) {
  vector<int> c;
  vector<int>::const_iterator it1 = a.begin(), it1e = a.end(), it2 = b.begin(),
                              it2e = b.end();
  while (c.size() < 10) {
    int a;
    if (it1 == it1e && it2 != it2e)
      a = *it2++;
    else if (it1 != it1e && it2 == it2e)
      a = *it1++;
    else if (it1 != it1e && it2 != it2e && *it1 < *it2)
      a = *it1++;
    else if (it1 != it1e && it2 != it2e)
      a = *it2++;
    else
      break;
    if (c.size() == 0 || c.back() != a) c.push_back(a);
  }
  return c;
}
void lcaBuild() {
  for (int n = 0; 1 << n <= N; n++)
    for (int i = 0; i < N; i++) {
      if (n) {
        P[i][n].t = P[P[i][n - 1].t][n - 1].t;
        P[i][n].prn = mergeVec(P[i][n - 1].prn, P[P[i][n - 1].t][n - 1].prn);
      } else {
        P[i][n].t = T[i];
        P[i][n].prn = mergeVec(nums[i], nums[T[i]]);
      }
    }
}
vector<int> lcaQuery(int p, int q) {
  if (L[p] < L[q]) swap(p, q);
  vector<int> ans;
  while (L[p] > L[q]) {
    auto& stuff = P[p][31 - __builtin_clz(L[p] - L[q])];
    ans = mergeVec(ans, stuff.prn);
    p = stuff.t;
  }
  if (p == q) return mergeVec(ans, nums[p]);
  for (int j = 31 - __builtin_clz(N); j >= 0; j--)
    if (P[p][j].t != P[q][j].t) {
      ans = mergeVec(ans, P[p][j].prn);
      ans = mergeVec(ans, P[q][j].prn);
      p = P[p][j].t;
      q = P[q][j].t;
    }
  ans = mergeVec(ans, P[p][0].prn);
  ans = mergeVec(ans, P[q][0].prn);
  return ans;
}
void dfs(int n, int p) {
  for (int m : adj[n])
    if (m != p) {
      T[m] = n;
      L[m] = L[n] + 1;
      dfs(m, n);
    }
}
int main() {
  int M, Q;
  scanf("%d%d%d", &N, &M, &Q);
  for (int n = 0; n < int(N - 1); n++) {
    int v, u;
    scanf("%d%d", &v, &u);
    v--;
    u--;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  for (int m = 0; m < int(M); m++) {
    int c;
    scanf("%d", &c);
    c--;
    if (nums[c].size() < 10) nums[c].push_back(m + 1);
  }
  dfs(0, -1);
  lcaBuild();
  for (int q = 0; q < int(Q); q++) {
    int v, u, a;
    scanf("%d%d%d", &v, &u, &a);
    v--;
    u--;
    auto ans = lcaQuery(v, u);
    int k = min((int)ans.size(), a);
    printf("%d", k);
    for (int i = 0; i < int(k); i++) printf(" %d", ans[i]);
    printf("\n");
  }
}