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;
vector<int> v[100010];
vector<int>::iterator it;
int N, M;
int L[100010];
int T[100010];
bool vis[100010] = {false};
vector<int> c[100010];
int p[20][100010];
vector<int> mnp[20][100010];
void dfs(int x) {
  int i, y;
  vis[x] = true;
  for (i = 0; i < v[x].size(); i++) {
    y = v[x][i];
    if (!vis[y]) {
      L[y] = L[x] + 1;
      T[y] = x;
      dfs(y);
    }
  }
}
void pre1() {
  int i, j, sz;
  for (i = 0; i < N; i++) {
    for (j = 0; (1 << j) < N; j++) p[j][i] = -1;
  }
  for (i = 1; i < N; i++) {
    p[0][i] = T[i];
  }
  for (i = 1; i < N; i++) mnp[0][i] = c[T[i]];
  for (j = 1; (1 << j) < N; j++) {
    for (i = 1; i < N; i++) {
      if (p[j - 1][i] != -1 && p[j - 1][p[j - 1][i]] != -1) {
        p[j][i] = p[j - 1][p[j - 1][i]];
        mnp[j][i].resize(mnp[j - 1][i].size() + mnp[j - 1][p[j - 1][i]].size());
        merge(mnp[j - 1][i].begin(), mnp[j - 1][i].end(),
              mnp[j - 1][p[j - 1][i]].begin(), mnp[j - 1][p[j - 1][i]].end(),
              mnp[j][i].begin());
        if (mnp[j][i].size() > 10) mnp[j][i].resize(10);
      }
    }
  }
}
vector<int> func(int u, int v) {
  int i;
  vector<int> p1;
  vector<int> p2;
  vector<int> path;
  vector<int> tmp;
  if (L[u] < L[v]) swap(u, v);
  p1 = c[u];
  p2 = c[v];
  int k = 31 - __builtin_clz(L[u]);
  if (L[u] != L[v]) {
    for (i = k; i >= 0; i--) {
      if (L[u] - (1 << i) >= L[v]) {
        tmp.resize(mnp[i][u].size() + p1.size());
        merge(mnp[i][u].begin(), mnp[i][u].end(), p1.begin(), p1.end(),
              tmp.begin());
        if (tmp.size() > 10) tmp.resize(10);
        p1 = tmp;
        u = p[i][u];
      }
    }
  }
  if (u == v) {
    path = p1;
    return path;
  }
  for (i = k; i >= 0; i--) {
    if (p[i][u] != -1 && p[i][u] != p[i][v]) {
      tmp.resize(mnp[i][u].size() + p1.size());
      merge(mnp[i][u].begin(), mnp[i][u].end(), p1.begin(), p1.end(),
            tmp.begin());
      if (tmp.size() > 10) tmp.resize(10);
      p1 = tmp;
      u = p[i][u];
      tmp.resize(mnp[i][v].size() + p2.size());
      merge(mnp[i][v].begin(), mnp[i][v].end(), p2.begin(), p2.end(),
            tmp.begin());
      if (tmp.size() > 10) tmp.resize(10);
      p2 = tmp;
      v = p[i][v];
    }
  }
  path.resize(p1.size() + p2.size());
  merge(p1.begin(), p1.end(), p2.begin(), p2.end(), path.begin());
  if (path.size() > 10) path.resize(10);
  tmp.resize(c[T[u]].size() + path.size());
  merge(c[T[u]].begin(), c[T[u]].end(), path.begin(), path.end(), tmp.begin());
  if (tmp.size() > 10) tmp.resize(10);
  path = tmp;
  return path;
}
int main() {
  int i, x, y, Q, z, M, a, no, k, j, sz;
  cin >> N >> M >> Q;
  for (i = 0; i < N - 1; i++) {
    cin >> x >> y;
    x--;
    y--;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 0; i < M; i++) {
    cin >> x;
    c[x - 1].push_back(i + 1);
  }
  for (i = 0; i < N; i++) {
    if (c[i].size() > 10) c[i].resize(10);
  }
  T[0] = -1;
  L[0] = 0;
  dfs(0);
  pre1();
  while (Q--) {
    int cnt = 0;
    cin >> x >> y >> a;
    vector<int> path;
    path = func(x - 1, y - 1);
    for (it = path.begin(); it != path.end(); it++) {
      if (*it > 0) cnt++;
    }
    k = min(cnt, a);
    cout << k << " ";
    for (it = path.begin(); it != path.end() && k > 0; it++, k--)
      cout << (*it) << " ";
    cout << endl;
  }
}