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 Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void err(istream_iterator<string> it) { cout << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ' ';
  err(++it, args...);
}
const int N = (int)2e5 + 5;
const int maxN = (int)1e6 + 6;
const long long Mod = (long long)1e9 + 7;
const int inf = (int)2e9;
const long long Inf = (long long)1e18;
struct lowest_common_ancestor {
  const static int N = 1e5 + 5;
  const static int k = 17;
  int par[N][k + 1], lev[N];
  vector<vector<int>> g, city;
  vector<int> ID[N][k + 1];
  lowest_common_ancestor() {
    for (int i = 0; i < N; ++i) {
      for (int j = 0; j <= k; ++j) par[i][j] = 0;
      lev[i] = 0;
    }
  }
  void merge(vector<int> &a, vector<int> &b) {
    vector<int> tmp;
    auto i = a.begin(), j = b.begin();
    for (int k = 1; k <= min(10, (int)a.size() + (int)b.size()); ++k) {
      if (i != a.end() and j != b.end()) {
        if (*i < *j)
          tmp.push_back(*i), ++i;
        else
          tmp.push_back(*j), ++j;
      } else if (i != a.end())
        tmp.push_back(*i), ++i;
      else
        tmp.push_back(*j), ++j;
    }
    a.assign(tmp.begin(), tmp.end());
  }
  void build(int s, int p = 0) {
    lev[s] = 1 + lev[p];
    par[s][0] = p;
    ID[s][0].assign(city[p].begin(), city[p].end());
    for (int i = 1; i <= k; ++i) {
      par[s][i] = par[par[s][i - 1]][i - 1];
      merge(ID[s][i], ID[s][i - 1]);
      merge(ID[s][i], ID[par[s][i - 1]][i - 1]);
    }
    for (int i : g[s])
      if (i != p) build(i, s);
  }
  void getLCA(int a, int b, int t) {
    if (lev[a] < lev[b]) swap(a, b);
    vector<int> res(city[a].begin(), city[a].end());
    int d = lev[a] - lev[b];
    int tt = b;
    for (int i = k; i >= 0; --i) {
      if ((d >> i) & 1) {
        merge(res, ID[a][i]);
        a = par[a][i];
      }
    }
    if (a == b) {
      printf("%d", min((int)res.size(), t));
      for (int i = 0; i < min((int)res.size(), t); ++i) printf(" %d", res[i]);
      cout << '\n';
      return;
    }
    for (int i = k; i >= 0; --i) {
      if (par[a][i] != par[b][i]) {
        merge(res, ID[a][i]);
        merge(res, ID[b][i]);
        b = par[b][i];
        a = par[a][i];
      }
    }
    merge(res, city[par[a][0]]);
    merge(res, city[tt]);
    printf("%d", min((int)res.size(), t));
    for (int i = 0; i < min((int)res.size(), t); ++i) printf(" %d", res[i]);
    cout << '\n';
    return;
  }
  void solve() {
    int n = Int(), m = Int(), q = Int();
    g.resize(n + 1);
    city.resize(n + 1);
    for (int i = 1; i < n; ++i) {
      int x = Int(), y = Int();
      g[x].emplace_back(y);
      g[y].emplace_back(x);
    }
    for (int i = 1; i <= m; ++i) {
      int c = Int();
      if ((int)city[c].size() < 10) city[c].emplace_back(i);
    }
    build(1);
    while (q--) {
      int l = Int(), r = Int(), t = Int();
      getLCA(l, r, t);
    }
  }
};
lowest_common_ancestor lca;
int main() {
  int test = 1, tc = 0;
  while (test--) {
    lca.solve();
  }
  return 0;
}