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 int MX = 100005, lg = 17;
int n, m, Q, L[MX], sp[lg + 1][MX];
vector<int> G[MX], V[lg + 1][MX];
void bfs() {
  queue<int> q;
  q.push(1), L[1] = 1, sp[0][1] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (auto v : G[u])
      if (!sp[0][v]) {
        q.push(v), L[v] = L[u] + 1, sp[0][v] = u;
      }
  }
  sp[0][1] = 0;
}
void marge(vector<int> v1, vector<int> v2, vector<int>& v) {
  v = v1;
  v.insert(v.end(), v2.begin(), v2.end());
  sort(v.begin(), v.end());
  while (v.size() > 10) {
    v.pop_back();
  }
}
void build_sptbl() {
  for (int i = 1; i <= lg; i++)
    for (int j = 1; j <= n; j++) {
      sp[i][j] = sp[i - 1][sp[i - 1][j]];
      marge(V[i - 1][j], V[i - 1][sp[i - 1][j]], V[i][j]);
    }
}
vector<int> query(int u, int v) {
  vector<int> ret;
  if (L[u] < L[v]) swap(u, v);
  for (int i = lg; i >= 0; i--) {
    if (L[sp[i][u]] >= L[v]) {
      marge(ret, V[i][u], ret);
      u = sp[i][u];
    }
  }
  for (int i = lg; i >= 0; i--) {
    if (sp[i][u] != sp[i][v]) {
      marge(ret, V[i][u], ret);
      marge(ret, V[i][v], ret);
      u = sp[i][u], v = sp[i][v];
    }
  }
  if (u != v) {
    marge(ret, V[0][u], ret);
    marge(ret, V[0][v], ret);
    u = sp[0][u], v = sp[0][v];
  }
  marge(ret, V[0][u], ret);
  return ret;
}
int main() {
  scanf("%d %d %d", &n, &m, &Q);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1, a; i <= m; i++) {
    scanf("%d", &a);
    V[0][a].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    vector<int> dmy;
    marge(dmy, V[0][i], V[0][i]);
  }
  bfs();
  build_sptbl();
  while (Q--) {
    int u, v, a;
    scanf("%d %d %d", &u, &v, &a);
    vector<int> ans = query(u, v);
    int sz = min(a, (int)ans.size());
    printf("%d", sz);
    for (int i = 0; i < sz; i++) {
      printf(" %d", ans[i]);
    }
    printf("\n");
  }
  return 0;
}