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 gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
vector<vector<int> > G;
int P[100005][17];
vector<int> city[100005][17];
vector<int> H;
void S(vector<int> &a) { sort(a.begin(), a.end()); }
void dbug(vector<int> a) {
  for (int i = 0; i < a.size(); ++i) printf("%d  ", a[i]);
  printf("\n");
}
vector<int> merge(const vector<int> &a, const vector<int> &b) {
  vector<int> c;
  int i = 0, j = 0;
  while (i < a.size() && j < b.size()) {
    if (a[i] == b[j]) {
      c.push_back(a[i]);
      i++, j++;
      continue;
    } else {
      if (a[i] < b[j])
        c.push_back(a[i++]);
      else
        c.push_back(b[j++]);
    }
    if (c.size() > 12) return c;
  }
  while (i < a.size()) {
    c.push_back(a[i++]);
    if (c.size() > 12) return c;
  }
  while (j < b.size()) {
    c.push_back(b[j++]);
    if (c.size() > 12) return c;
  }
  return c;
}
void dfs(int u, int p = -1) {
  P[u][0] = p;
  for (int i = 1; i < 17; ++i) {
    if (P[u][i - 1] != -1) {
      P[u][i] = P[P[u][i - 1]][i - 1];
      city[u][i] = merge(city[u][i - 1], city[P[u][i - 1]][i - 1]);
    }
  }
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    if (v != p) {
      H[v] = H[u] + 1;
      dfs(v, u);
    }
  }
}
void swap(int &a, int &b) {
  int t = a;
  a = b;
  b = t;
}
int getp(int v, int k) {
  for (int i = 0; i < 17; ++i)
    if ((1 << i) & k) v = P[v][i];
  return v;
}
vector<int> get(int v, int k) {
  vector<int> ret;
  for (int i = 0; i < 17; ++i)
    if (P[v][i] != -1 && ((1 << i) & k)) {
      ret = merge(ret, city[v][i]);
      v = P[v][i];
    }
  return ret;
}
int lca(int v, int u) {
  vector<int> res;
  if (H[v] < H[u]) swap(u, v);
  v = getp(v, H[v] - H[u]);
  if (v == u) return v;
  for (int i = 17 - 1; i >= 0; --i) {
    if (P[v][i] != P[u][i]) {
      v = P[v][i];
      u = P[u][i];
    }
  }
  return P[v][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, q;
  cin >> n >> m >> q;
  G.assign(n + 1, vector<int>());
  H.assign(n + 1, 0);
  memset((P), (-1), sizeof(P));
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 0; i < m; ++i) {
    int u;
    cin >> u;
    vector<int> tmp;
    tmp.push_back(i);
    city[u][0] = merge(city[u][0], tmp);
  }
  dfs(1, 0);
  for (int i = 0; i < q; ++i) {
    int src, tar, a;
    cin >> src >> tar >> a;
    int p = lca(src, tar);
    vector<int> ans1 = get(src, H[src] - H[p]);
    vector<int> ans2 = get(tar, H[tar] - H[p] + 1);
    vector<int> ans = merge(ans1, ans2);
    cout << min((int)ans.size(), a);
    if (ans.size()) {
      cout << " ";
      cout << ans[0] + 1;
      for (int i = 1; i < ans.size() && i < a; ++i) cout << " " << (ans[i] + 1);
    }
    cout << endl;
  }
  return 0;
}