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>
#include <bits/stdc++.h>
using namespace std;
void solve();
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
struct stree;
const int N = 100001, M = 18, K = 10;
int n, m, q;
vector<int> g[N];
vector<int> c[N];
bool used[N];
int tin[N], tout[N], tctr;
int p[N][M], s[N], idx[N], bottom[N], top[N], len[N];
stree* st[N];
void dfs(int u, int par = -1) {
  used[u] = true;
  s[u] = 1;
  tin[u] = tctr++;
  p[u][0] = (par == -1 ? u : par);
  for (int i = 1; i < M; i++) p[u][i] = p[p[u][i - 1]][i - 1];
  for (int v : g[u])
    if (v != par) {
      dfs(v, u);
      s[u] += s[v];
    }
  idx[u] = 0, bottom[u] = u;
  for (int v : g[u])
    if (v != par)
      if (s[v] > (s[u] - 1) / 2) {
        idx[u] = idx[v] + 1;
        bottom[u] = bottom[v];
      }
  tout[u] = tctr++;
}
bool is_par(int u, int v) { return tin[u] <= tin[v] && tout[u] >= tout[v]; }
int lca(int u, int v) {
  if (is_par(u, v)) return u;
  if (is_par(v, u)) return v;
  for (int i = M - 1; i >= 0; i--)
    if (!is_par(p[u][i], v)) u = p[u][i];
  return p[u][0];
}
void mymerge(int k, const vector<int>& a, const vector<int>& b,
             vector<int>& res) {
  int n1 = a.size(), n2 = b.size();
  res.resize(n1 + n2);
  auto it = set_union(a.begin(), a.end(), b.begin(), b.end(), res.begin());
  k = min(k, (int)(it - res.begin()));
  res.resize(k);
}
struct stree {
  stree(int u, int len) {
    sz = 1;
    while (sz < len) sz <<= 1;
    st = new vector<int>[2 * sz];
    for (int i = 0; i < len; i++) {
      for (int j = 0; j < min(K, (int)c[u].size()); j++)
        st[sz + i].push_back(c[u][j]);
      u = p[u][0];
    }
    for (int i = sz - 1; i > 0; i--) {
      mymerge(K, st[2 * i], st[2 * i + 1], st[i]);
    }
  }
  vector<int> query(int l, int r, int L = 0, int R = -1, int k = 1) {
    if (R == -1) R = sz;
    if (l == L && r == R) return st[k];
    int M = L + (R - L) / 2;
    if (r <= M) return query(l, r, L, M, 2 * k);
    if (l >= M) return query(l, r, M, R, 2 * k + 1);
    vector<int> lret = query(l, M, L, M, 2 * k);
    vector<int> rret = query(M, r, M, R, 2 * k + 1);
    vector<int> ret;
    mymerge(K, lret, rret, ret);
    return ret;
  }
  int sz;
  vector<int>* st;
};
vector<int> query(int u, int v) {
  assert(is_par(u, v));
  vector<int> ret;
  while (true) {
    int c = bottom[v];
    int w = (bottom[u] == c ? u : top[c]);
    vector<int> stres = st[c]->query(idx[v], idx[w] + 1);
    vector<int> tmp;
    mymerge(K, ret, stres, tmp);
    ret.swap(tmp);
    v = top[c];
    if (v == 1 || is_par(v, u)) break;
    v = p[v][0];
  }
  return ret;
}
void solve() {
  while (cin >> n >> m >> q) {
    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 x;
      cin >> x;
      c[x].push_back(i + 1);
    }
    dfs(1);
    for (int i = 1; i <= n; i++) {
      int j = bottom[i];
      if (idx[i] + 1 > len[j]) {
        len[j] = idx[i] + 1;
        top[j] = i;
      }
    }
    for (int i = 1; i <= n; i++) {
      sort(c[i].begin(), c[i].end());
      if (bottom[i] == i) st[i] = new stree(i, len[i]);
    }
    for (int i = 0; i < q; i++) {
      int u, v, k;
      cin >> u >> v >> k;
      int p = lca(u, v);
      vector<int> up = query(p, u);
      vector<int> vp = query(p, v);
      vector<int> ans;
      mymerge(k, up, vp, ans);
      cout << ans.size() << " ";
      for (int x : ans) cout << x << " ";
      cout << endl;
    }
  }
}
