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 N = 1e5 + 5, L = 20;
int n, m, q;
vector<int> adj[N];
vector<int> C[N][L];
int to[N][L];
int dep[N];
void dfs(int u, int p) {
  to[u][0] = p;
  dep[u] = 1 + dep[p];
  for (auto v : adj[u])
    if (v != p) dfs(v, u);
}
vector<int> relax(vector<int> v) {
  vector<int> ret;
  ret.clear();
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size() && i < 10; i++) ret.push_back(v[i]);
  return ret;
}
vector<int> mrg(vector<int> a, vector<int> b) {
  for (int B : b) a.push_back(B);
  return relax(a);
}
vector<int> query(int u, int v) {
  if (u == v) {
    return C[u][0];
  }
  if (dep[u] < dep[v]) swap(u, v);
  vector<int> ret;
  ret.clear();
  int k = dep[u] - dep[v];
  for (int i = 0; i < L; i++)
    if (k & (1 << i)) {
      ret = mrg(ret, C[u][i]);
      u = to[u][i];
    }
  if (u == v) {
    ret = mrg(ret, C[u][0]);
    return ret;
  }
  for (int i = L - 1; i >= 0; i--) {
    if (to[u][i] != to[v][i]) {
      ret = mrg(ret, C[u][i]);
      ret = mrg(ret, C[v][i]);
      u = to[u][i];
      v = to[v][i];
    }
  }
  ret = mrg(ret, C[u][0]);
  ret = mrg(ret, C[v][0]);
  u = to[u][0];
  ret = mrg(ret, C[u][0]);
  return ret;
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &u);
    C[u][0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    C[i][0] = relax(C[i][0]);
  }
  dfs(1, 0);
  for (int j = 1; j < L; j++) {
    for (int i = 1; i <= n; i++) {
      to[i][j] = to[to[i][j - 1]][j - 1];
      C[i][j] = mrg(C[i][j - 1], C[to[i][j - 1]][j - 1]);
    }
  }
  int a;
  while (q--) {
    scanf("%d %d %d", &u, &v, &a);
    vector<int> ans = query(u, v);
    printf("%d ", min(int(ans.size()), a));
    for (int i = 0; i < a && i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
}