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, i, j, T, level, subtree[100111], centroid[100111][20], nc, M, Q, ans[20];
vector<vector<int>> G;
vector<int> persons[100111], plist[100111][20];
set<int> ps;
bool vis[100111];
void dfs(int v, int p);
int find_centroid(int v, int sz);
void dfs2(int v, int p);
int main() {
  scanf("%d", &N);
  scanf("%d", &M);
  scanf("%d", &Q);
  G.resize(N + 1);
  for (i = 0; i < (N - 1); i++) {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (i = 0; i < (M); i++) {
    int a;
    scanf("%d", &a);
    persons[a].push_back(i + 1);
  }
  queue<pair<int, int>> q;
  q.push({1, 1});
  while (!q.empty()) {
    pair<int, int> c = q.front();
    q.pop();
    level = c.second;
    dfs(c.first, -1);
    nc = find_centroid(c.first, subtree[c.first]);
    dfs2(nc, -1);
    vis[nc] = true;
    for (auto &v : G[nc])
      if (!vis[v]) q.push({v, level + 1});
  }
  while (Q--) {
    int v, u, a, lv, lu;
    scanf("%d", &v);
    scanf("%d", &u);
    scanf("%d", &a);
    for (lv = 1; centroid[v][lv]; lv++)
      ;
    lv--;
    for (lu = 1; centroid[u][lu]; lu++)
      ;
    lu--;
    int lca = min(lv, lu);
    while (centroid[v][lca] != centroid[u][lca]) lca--;
    int taken = 0;
    vector<int> &first = plist[v][lca], &second = plist[u][lca];
    auto f = first.begin(), s = second.begin();
    while (taken < a && (f != first.end() || s != second.end())) {
      if (f == first.end()) {
        ans[taken++] = *s;
        s++;
      } else if (s == second.end()) {
        ans[taken++] = *f;
        f++;
      } else if (*f < *s) {
        ans[taken++] = *f;
        f++;
      } else if (*f > *s) {
        ans[taken++] = *s;
        s++;
      } else {
        ans[taken++] = *f;
        f++;
        s++;
      }
    }
    printf("%d", taken);
    for (i = 0; i < (taken); i++) printf(" %d", ans[i]);
    printf("\n");
  }
}
void dfs(int v, int p) {
  subtree[v] = 1;
  for (auto &n : G[v]) {
    if (!vis[n] && n != p) {
      dfs(n, v);
      subtree[v] += subtree[n];
    }
  }
}
int find_centroid(int v, int sz) {
  int branch = v;
  for (auto &n : G[v]) {
    if (!vis[n] && subtree[n] > sz / 2) branch = n;
  }
  if (branch == v) return v;
  subtree[v] -= subtree[branch];
  subtree[branch] += subtree[v];
  return find_centroid(branch, sz);
}
void dfs2(int v, int p) {
  centroid[v][level] = nc;
  ps.insert((persons[v]).begin(), (persons[v]).end());
  auto it = ps.begin();
  for (int i = 0; i < 10 && it != ps.end(); i++, it++)
    plist[v][level].push_back(*it);
  for (auto &n : G[v]) {
    if (!vis[n] && n != p) dfs2(n, v);
  }
  for (auto x : persons[v]) ps.erase(x);
}