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;
void merge(const vector<int> &v1, const vector<int> &v2, vector<int> &v) {
  int i, j;
  for (i = 0, j = 0; i < (int)v1.size() && j < (int)v2.size();) {
    if (v1[i] < v2[j]) {
      v.push_back(v1[i]);
      i++;
    } else if (v1[i] > v2[j]) {
      v.push_back(v2[j]);
      j++;
    } else {
      i++;
    }
    if (v.size() >= 10) return;
  }
  while (i < (int)v1.size() && v.size() < 10) v.push_back(v1[i++]);
  while (j < (int)v2.size() && v.size() < 10) v.push_back(v2[j++]);
}
int n, m, q;
vector<int> g[100005];
int par[100005][25];
int dep[100005];
vector<int> person[100005];
vector<int> v[100005][25];
int id[100005];
void dfs(int u, int p) {
  dep[u] = dep[p] + 1;
  par[u][0] = p;
  merge(person[p], person[u], v[u][0]);
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (v == p) continue;
    dfs(v, u);
  }
}
void solve(int a, int b, vector<int> &vec) {
  if (a == b) vec = person[a];
  vector<int> tmp;
  if (dep[a] > dep[b]) swap(a, b);
  for (int i = 20; i >= 0; i--) {
    if ((dep[b] - dep[a]) & (1 << i)) {
      tmp.clear();
      merge(v[b][i], vec, tmp);
      vec = tmp;
      b = par[b][i];
    }
  }
  if (a == b) return;
  for (int i = 20; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      tmp.clear();
      merge(v[b][i], vec, tmp);
      vec = tmp;
      tmp.clear();
      merge(v[a][i], vec, tmp);
      vec = tmp;
      a = par[a][i];
      b = par[b][i];
    }
  }
  tmp.clear();
  merge(v[b][0], vec, tmp);
  vec = tmp;
  tmp.clear();
  merge(v[a][0], vec, tmp);
  vec = tmp;
  a = par[a][0];
  b = par[b][0];
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &id[i]);
    person[id[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) sort(person[i].begin(), person[i].end());
  dfs(1, 1);
  for (int j = 1; j <= 20; j++) {
    for (int i = 1; i <= n; i++) {
      par[i][j] = par[par[i][j - 1]][j - 1];
      merge(v[i][j - 1], v[par[i][j - 1]][j - 1], v[i][j]);
    }
  }
  for (int i = 1; i <= q; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    vector<int> vec;
    solve(a, b, vec);
    int len = min((int)vec.size(), c);
    printf("%d ", len);
    for (int i = 0; i < len; i++) printf("%d ", vec[i]);
    printf("\n");
  }
  return 0;
}