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, m, q;
vector<int> G[100007], H[100007];
int V[100007][11] = {0}, dep[100007] = {0}, fa[100007], top[100007];
bool vis[100007] = {0}, hvy[100007] = {0};
int A[100007] = {0}, ofs[100007], A_cnt = 0;
int T[100007 << 4][11];
int dfs(int u, int p, int d) {
  dep[u] = d, fa[u] = p;
  int size = 1, maxs = 0, maxv = 0;
  for (auto it = G[u].begin(); it != G[u].end(); it++)
    if (!dep[*it]) {
      int res = dfs(*it, u, d + 1);
      if (res > maxs) maxs = res, maxv = *it;
      size += res;
    }
  if (maxs) hvy[maxv] = true;
  return size;
}
void dfs2(int u) {
  vis[u] = true;
  if (hvy[u])
    top[u] = top[fa[u]];
  else
    top[u] = u;
  H[top[u]].push_back(u);
  for (auto it = G[u].begin(); it != G[u].end(); it++)
    if (!vis[*it]) dfs2(*it);
}
void merge(int *s1, int *s2, int *tar) {
  tar[0] = 0;
  int i = 1, j = 1;
  while (tar[0] < 10 && i <= s1[0] && j <= s2[0]) {
    if (s1[i] < s2[j])
      tar[++tar[0]] = s1[i++];
    else
      tar[++tar[0]] = s2[j++];
  }
  while (tar[0] < 10 && i <= s1[0]) tar[++tar[0]] = s1[i++];
  while (tar[0] < 10 && j <= s2[0]) tar[++tar[0]] = s2[j++];
}
void build(int root, int s, int e) {
  if (s == e) {
    T[root][0] = 0;
    for (int i = 1; i <= V[A[s]][0]; i++) T[root][++T[root][0]] = V[A[s]][i];
    return;
  }
  int mid = (s + e) >> 1;
  int l = (root << 1), r = 1 + (root << 1);
  build(l, s, mid);
  build(r, mid + 1, e);
  merge(T[l], T[r], T[root]);
}
void query_(int root, int s, int e, int qs, int qe, int *res) {
  int tmp[11];
  tmp[0] = 0;
  if (qe < s || qs > e) return;
  if (qs <= s && e <= qe) {
    merge(res, T[root], tmp);
    for (int i = 0; i < 11; i++) res[i] = tmp[i];
    return;
  }
  int mid = (s + e) >> 1;
  int l = (root << 1), r = 1 + (root << 1);
  query_(l, s, mid, qs, qe, res);
  query_(r, mid + 1, e, qs, qe, res);
}
void query(int qs, int qe, int *res) {
  res[0] = 0;
  if (qs > qe) return;
  query_(1, 0, A_cnt - 1, qs, qe, res);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &u);
    if (V[u][0] < 10) V[u][++V[u][0]] = i;
  }
  for (int i = 1; i <= n; i++) sort(V[i] + 1, V[i] + V[i][0] + 1);
  dfs(1, 0, 1);
  dfs2(1);
  int end = 0;
  for (int i = 1; i <= n; i++) {
    for (auto it = H[i].begin(); it != H[i].end(); it++) {
      A[A_cnt++] = *it;
      ofs[top[*it]] = end;
    }
    end += H[i].size();
  }
  build(1, 0, A_cnt - 1);
  int cmd, a;
  int ans[11], res[11], tmp[11];
  for (int i = 0; i < q; i++) {
    scanf("%d%d%d", &u, &v, &a);
    ans[0] = 0;
    while (1) {
      int z;
      if (top[u] == top[v]) break;
      if (dep[top[u]] > dep[top[v]])
        u = fa[top[z = u]];
      else
        v = fa[top[z = v]];
      query(ofs[top[z]], ofs[top[z]] + dep[z] - dep[top[z]], res);
      merge(ans, res, tmp);
      for (int i = 0; i < 11; i++) ans[i] = tmp[i];
    }
    if (dep[u] > dep[v]) swap(u, v);
    query(ofs[top[u]] + dep[u] - dep[top[u]],
          ofs[top[u]] + dep[v] - dep[top[u]], res);
    merge(ans, res, tmp);
    for (int i = 0; i < 11; i++) ans[i] = tmp[i];
    ans[0] = min(ans[0], a);
    for (int i = 0; i <= ans[0]; i++) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}