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 MAX_N = 100005;
const int MAX_M = MAX_N * 2;
const int K = 18;
int head[MAX_N], pnt[MAX_M], nt[MAX_M], fa[MAX_N][K], d[MAX_N], ptr = 0;
vector<int> cnt[MAX_N];
vector<int> fn[MAX_N][K];
inline void addedge(int x, int y) {
  pnt[ptr] = y;
  nt[ptr] = head[x];
  head[x] = ptr++;
  pnt[ptr] = x;
  nt[ptr] = head[y];
  head[y] = ptr++;
}
inline void init() {
  ptr = 0;
  memset(head, -1, sizeof head);
}
inline void merge(vector<int>& v1, const vector<int>& v2) {
  for (int i = 0; i < (int)v2.size(); ++i) v1.push_back(v2[i]);
  sort(v1.begin(), v1.end());
  int len = unique(v1.begin(), v1.end()) - v1.begin();
  v1.resize(min(len, 10));
}
void dfs(int cur, int p, int dep) {
  fa[cur][0] = p;
  d[cur] = dep;
  fn[cur][0] = cnt[cur];
  merge(fn[cur][0], cnt[p]);
  for (int i = 1; i < K; ++i) {
    fa[cur][i] = fa[fa[cur][i - 1]][i - 1];
    fn[cur][i] = fn[cur][i - 1];
    merge(fn[cur][i], fn[fa[cur][i - 1]][i - 1]);
  }
  for (int i = head[cur]; i != -1; i = nt[i]) {
    int to = pnt[i];
    if (to != p) {
      dfs(to, cur, dep + 1);
    }
  }
}
vector<int> res;
inline void query(int u, int v, int a) {
  res.clear();
  if (d[u] < d[v]) swap(u, v);
  int diff = d[u] - d[v];
  for (int i = 0; diff && i < K; ++i) {
    if (diff & (1 << i)) {
      merge(res, fn[u][i]);
      diff -= (1 << i);
      u = fa[u][i];
    }
  }
  if (u == v) {
    merge(res, cnt[u]);
    if ((int)res.size() > a) res.resize(a);
    printf("%d ", (int)res.size());
    for (int i = 0; i < (int)res.size(); ++i) printf("%d ", res[i]);
    printf("\n");
    return;
  }
  for (int i = K - 1; i >= 0; --i) {
    if (fa[u][i] != fa[v][i]) {
      merge(res, fn[u][i]);
      merge(res, fn[v][i]);
      u = fa[u][i], v = fa[v][i];
    }
  }
  merge(res, fn[u][0]);
  merge(res, fn[v][0]);
  if ((int)res.size() > a) res.resize(a);
  printf("%d ", (int)res.size());
  for (int i = 0; i < (int)res.size(); ++i) printf("%d ", res[i]);
  printf("\n");
}
int main() {
  init();
  int n, m, q, u, v, c, a;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < (int)n - 1; ++i) {
    scanf("%d %d", &u, &v);
    addedge(u, v);
  }
  for (int i = 0; i < (int)m; ++i) {
    scanf("%d", &c);
    cnt[c].push_back(i + 1);
  }
  for (int i = 0; i < (int)n; ++i) {
    sort(cnt[i + 1].begin(), cnt[i + 1].end());
    if (cnt[i + 1].size() > 10) {
      cnt[i + 1].resize(10);
    }
  }
  dfs(1, 1, 0);
  for (int i = 0; i < (int)q; ++i) {
    scanf("%d %d %d", &u, &v, &a);
    query(u, v, a);
  }
  return 0;
}