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;
int d[123456], par[123456][30];
vector<int> G[123456], val[123456][30];
vector<int> operator+(vector<int> &v, vector<int> &u) {
  vector<int> re;
  unsigned int i = 0, j = 0;
  while (i < v.size() && j < u.size()) {
    if (v[i] < u[j])
      re.push_back(v[i++]);
    else
      re.push_back(u[j++]);
    if (re.size() >= 10) return re;
  }
  while (i < v.size()) {
    re.push_back(v[i++]);
    if (re.size() >= 10) return re;
  }
  while (j < u.size()) {
    re.push_back(u[j++]);
    if (re.size() >= 10) return re;
  }
  return re;
}
vector<int> operator+(vector<int> &v, int x) {
  vector<int> re;
  unsigned int i = 0;
  while (i < v.size() && v[i] < x && re.size() < 10) re.push_back(v[i++]);
  if (re.size() < 10) re.push_back(x);
  while (re.size() < 10 && i < v.size()) re.push_back(v[i++]);
  return re;
}
void addedge(int x, int y) { G[x].push_back(y); }
void dfs(int u, int p) {
  par[u][0] = p;
  for (int j = 1; j < 30; j++) {
    par[u][j] = par[par[u][j - 1]][j - 1];
    val[u][j] = val[u][j - 1] + val[par[u][j - 1]][j - 1];
  }
  for (unsigned int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != p) {
      d[v] = d[u] + 1;
      dfs(v, u);
    }
  }
}
int get_par(int u, int k) {
  for (int j = 0; j < 30; j++)
    if ((1 << j) & k) u = par[u][j];
  return u;
}
int lca(int u, int v) {
  if (d[u] < d[v]) swap(u, v);
  u = get_par(u, d[u] - d[v]);
  if (u == v) return u;
  for (int j = 30 - 1; j >= 0; j--)
    if (par[u][j] != par[v][j]) {
      u = par[u][j];
      v = par[v][j];
    }
  return par[u][0];
}
vector<int> get(int u, int k) {
  vector<int> re;
  for (int j = 0; j < 30; j++)
    if ((1 << j) & k) {
      re = re + val[u][j];
      u = par[u][j];
    }
  return re;
}
void init() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    addedge(x, y);
    addedge(y, x);
  }
  for (int i = 1; i <= m; i++) {
    int tmp;
    scanf("%d", &tmp);
    val[tmp][0] = val[tmp][0] + i;
  }
  dfs(1, 0);
}
void solve() {
  for (int i = 1; i <= q; i++) {
    int u, v, k, p;
    scanf("%d%d%d", &u, &v, &k);
    p = lca(u, v);
    vector<int> tmp1 = get(u, d[u] - d[p]), tmp2 = get(v, d[v] - d[p] + 1);
    vector<int> ans = tmp1 + tmp2;
    if (k > ans.size()) k = ans.size();
    printf("%d", k);
    for (int i = 0; i < k; i++) printf(" %d", ans[i]);
    printf("\n");
  }
}
int main() {
  init();
  solve();
}