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;
long long MOD = 1000000007;
vector<std::vector<int> > List(100000);
vector<std::vector<int> > id(100000);
int parent[18][100000];
int level[100000];
int m[18][100000][10];
int ans[20];
void merge(int v, int l, int u) {
  int i = 0, j = 0;
  assert(l > 0);
  for (int k = 0; k < 10; k++) {
    if (m[l - 1][v][i] < m[l - 1][u][j]) {
      m[l][v][k] = m[l - 1][v][i];
      i++;
    } else {
      m[l][v][k] = m[l - 1][u][j];
      j++;
    }
  }
}
void DFS(int root, bool visited[]) {
  visited[root] = true;
  for (int i = 0; i < (int)List[root].size(); ++i) {
    int v = List[root][i];
    if (!visited[v]) {
      parent[0][v] = root;
      sort(id[v].begin(), id[v].end());
      for (int j = 0; j < (int)id[v].size() and j < 10; j++)
        m[0][v][j] = id[v][j];
      bool rootDone = false;
      for (int j = 1; j < 18; ++j) {
        parent[j][v] = parent[j - 1][parent[j - 1][v]];
        if (rootDone) {
          assert(j > 0);
          for (int k = 0; k < 10; k++) m[j][v][k] = m[j - 1][v][k];
          continue;
        }
        if (parent[j - 1][v] == 0) rootDone = true;
        merge(v, j, parent[j - 1][v]);
      }
      level[v] = level[root] + 1;
      DFS(v, visited);
    }
  }
  return;
}
int tmp[20];
void mergeInto(int v, int l) {
  for (int i = 0; i < 10; i++) tmp[i] = ans[i];
  int i = 0, j = 0;
  for (int k = 0; k < 10; k++) {
    if (tmp[i] < m[l][v][j]) {
      ans[k] = tmp[i];
      i++;
    } else {
      ans[k] = m[l][v][j];
      j++;
    }
  }
}
int LCA(int x, int y) {
  if (level[x] > level[y]) swap(x, y);
  int diff = level[y] - level[x];
  for (int i = 0; i < 10; i++) ans[i] = MOD;
  while (diff) {
    int p = (diff & -diff);
    mergeInto(y, (int)log2(p));
    y = parent[(int)log2(p)][y];
    diff -= p;
  }
  if (x == y) {
    mergeInto(x, 0);
    return x;
  }
  for (int i = 17; i >= 0; --i) {
    if (parent[i][x] != parent[i][y]) {
      mergeInto(x, i);
      mergeInto(y, i);
      x = parent[i][x], y = parent[i][y];
    }
  }
  for (int i = 0; i <= 17; i++) {
    if (parent[i][x] != parent[i][y]) {
      mergeInto(x, i);
      mergeInto(y, i);
      x = parent[i][x], y = parent[i][y];
    }
  }
  mergeInto(parent[0][x], 0);
  mergeInto(x, 0);
  mergeInto(y, 0);
  return parent[0][x];
}
void initialize_LCA(int n) {
  bool vis[100000];
  memset(vis, 0, sizeof vis);
  for (int i = 0; i < 18; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < 10; k++) {
        m[i][j][k] = MOD;
      }
    }
  }
  sort(id[0].begin(), id[0].end());
  for (int i = 0; i < 18; ++i) {
    parent[i][0] = 0;
    for (int j = 0; j < id[0].size(); j++) m[i][0][j] = id[0][j];
  }
  level[0] = 0;
  DFS(0, vis);
}
int main() {
  int n, M, q;
  scanf("%d%d%d", &n, &M, &q);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--, b--;
    List[a].push_back(b);
    List[b].push_back(a);
  }
  for (int i = 1; i <= M; i++) {
    int c;
    scanf("%d", &c);
    c--;
    id[c].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < id[i].size(); j++) {
    }
  }
  initialize_LCA(n);
  std::vector<int> out;
  for (int i = 0; i < n; i++) {
    for (int k = 0; k < 1; k++) {
    }
  }
  while (q--) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    u--, v--;
    LCA(u, v);
    out.clear();
    for (int i = 0; i < a; i++) {
      if (ans[i] == MOD) break;
      out.push_back(ans[i]);
    }
    printf("%d ", (int)out.size());
    for (int i = 0; i < out.size(); i++) {
      printf("%d ", out[i]);
    }
    printf("\n");
  }
}