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 INF = 1e9 + 7, mod = 1e9 + 9, pw = 31;
const double eps = 1e-11;
int n, m, q;
vector<int> a[105000], d[105000][20];
vector<int> g[105000];
int in[105000], out[105000];
bool used[105000];
int up[105000][20], l = 18;
vector<int> res;
vector<int> add(vector<int> s, vector<int> t) {
  res.resize(0);
  int i = 0, j = 0;
  while (res.size() <= 10) {
    if (i >= s.size() && j >= t.size()) break;
    if (i < s.size() && j < t.size() && s[i] == t[j]) {
      res.push_back(s[i++]);
      while (j < t.size() && t[j] == s[i - 1]) j++;
      while (i < s.size() && s[i] == s[i - 1]) i++;
    } else if (i < s.size() && (j >= t.size() || s[i] < t[j]))
      res.push_back(s[i++]);
    else
      res.push_back(t[j++]);
  }
  return res;
}
void dfs(int v = 1, int p = 1) {
  static int kol = 0;
  in[v] = ++kol;
  used[v] = 1;
  up[v][0] = p;
  d[v][0] = add(a[v], a[p]);
  for (int i = 1; i <= l; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    d[v][i] = add(d[v][i - 1], d[up[v][i - 1]][i - 1]);
  }
  for (int i = 0; i < g[v].size(); i++)
    if (!used[g[v][i]]) dfs(g[v][i], v);
  out[v] = ++kol;
}
bool ok(int u, int v) {
  if (in[u] <= in[v] && out[u] >= out[v]) return 1;
  return 0;
}
int lca(int a, int b) {
  if (ok(a, b)) return a;
  if (ok(b, a)) return b;
  for (int i = l; i >= 0; --i)
    if (!ok(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
vector<int> find(int a, int b) {
  res.resize(0);
  for (int i = l; i >= 0; --i)
    if (ok(b, up[a][i])) {
      res = add(res, d[a][i]);
      a = up[a][i];
    }
  return res;
}
vector<int> get(int u, int v) {
  res.resize(0);
  int to = lca(u, v);
  res = add(find(u, to), find(v, to));
  res = add(res, a[u]);
  res = add(res, a[v]);
  res = add(res, a[to]);
  return res;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    a[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(a[i].begin(), a[i].end());
    a[i].resize(min((int)a[i].size(), 10));
  }
  dfs();
  vector<int> s;
  for (int i = 1; i <= q; i++) {
    int u, v, x;
    scanf("%d%d%d", &u, &v, &x);
    s = get(u, v);
    printf("%d ", min((int)s.size(), x));
    for (int i = 0; i < x && i < s.size(); i++) printf("%d ", s[i]);
    printf("\n");
  }
  return 0;
}