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;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
T abs(T x) {
  return x < 0 ? -x : x;
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
const int MAXN = 1e5 + 10;
int n, m, k, l, t;
vector<int> d[MAXN], people[MAXN], human[MAXN][27];
int p[MAXN][27];
int tin[MAXN], tout[MAXN];
vector<int> merge_unic(vector<int> a, const vector<int> &b) {
  set<int> s;
  for (int i = 0; i < b.size(); i++) s.insert(b[i]);
  for (int i = 0; i < a.size(); i++) s.insert(a[i]);
  a.clear();
  int i = 1;
  for (set<int>::iterator it = s.begin(); it != s.end() && i <= 10; it++, i++)
    a.push_back(*it);
  return a;
}
vector<int> merge(vector<int> a, const vector<int> &b) {
  for (int i = 0; i < b.size(); i++) a.push_back(b[i]);
  sort(a.begin(), a.end());
  while (a.size() > 10) a.pop_back();
  return a;
}
void dfs(int x, int pr = 0) {
  tin[x] = ++t;
  p[x][0] = pr;
  human[x][0] = people[pr];
  for (int i = 1; i <= l; i++)
    p[x][i] = p[p[x][i - 1]][i - 1],
    human[x][i] = merge(human[x][i - 1], human[p[x][i - 1]][i - 1]);
  for (int i = 0; i < d[x].size(); i++)
    if (d[x][i] != pr) dfs(d[x][i], x);
  tout[x] = ++t;
}
bool f(const int &x, const int &y) {
  return tin[x] <= tin[y] && tout[x] >= tout[y];
}
int lca(int a, const int &b) {
  if (f(a, b)) return a;
  if (f(b, a)) return b;
  for (int i = l; i >= 0; i--)
    if (!f(p[a][i], b) && p[a][i]) a = p[a][i];
  return p[a][0] ? p[a][0] : a;
}
vector<int> back(int a, const int &b) {
  vector<int> res;
  res.clear();
  for (int i = l; i >= 0 && a != b; i--)
    if (tout[b] >= tout[p[a][i]]) res = merge(res, human[a][i]), a = p[a][i];
  return res;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    d[x].push_back(y);
    d[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    people[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(people[i].begin(), people[i].end());
    while (people[i].size() > 10) people[i].pop_back();
  }
  l = 1;
  while ((1 << l) <= n) l++;
  dfs(1);
  tout[0] = ++t;
  while (k--) {
    int x, y, a, z;
    vector<int> res;
    cin >> x >> y >> a;
    z = lca(x, y);
    if (x == y)
      res = people[x];
    else if (x == z) {
      res = merge(people[y], back(y, z));
    } else if (y == z)
      res = merge(people[x], back(x, z));
    else
      res = merge(merge(people[x], people[y]),
                  merge_unic(back(x, z), back(y, z)));
    cout << min((int)res.size(), a) << " ";
    for (int i = 0; i < min((int)res.size(), a); i++) cout << res[i] << " ";
    cout << endl;
  }
}