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;
struct node {
  int l, r, sum;
  node() {}
  node(int a, int b, int c) {
    l = a;
    r = b;
    sum = c;
  }
};
node st[10101010];
vector<int> g[101010], lives[101010];
int lca[101010][20], h[101010], root[101010], parent[101010];
int cnt = 1;
int n, m, q;
int update(int no, int l, int r, int pos, int val) {
  if (pos < l || pos > r) return no;
  int clone = cnt++;
  st[clone] = st[no];
  if (l == r) {
    st[clone].sum += val;
    return clone;
  }
  int mid = (l + r) / 2;
  st[clone].l = update(st[clone].l, l, mid, pos, val);
  st[clone].r = update(st[clone].r, mid + 1, r, pos, val);
  st[clone].sum = st[st[clone].l].sum + st[st[clone].r].sum;
  return clone;
}
int query(int noU, int noV, int noLCA, int noPai, int l, int r, int k) {
  if (l == r) {
    int sum = st[noU].sum + st[noV].sum - st[noLCA].sum - st[noPai].sum;
    if (k == 1 && sum == 1) return l;
    return -1;
  }
  int sumEsq = st[st[noU].l].sum + st[st[noV].l].sum - st[st[noLCA].l].sum -
               st[st[noPai].l].sum;
  int mid = (l + r) / 2;
  if (sumEsq >= k)
    return query(st[noU].l, st[noV].l, st[noLCA].l, st[noPai].l, l, mid, k);
  return query(st[noU].r, st[noV].r, st[noLCA].r, st[noPai].r, mid + 1, r,
               k - sumEsq);
}
void dfs(int u, int pai) {
  if (u == 1)
    lca[u][0] = u;
  else
    lca[u][0] = pai;
  h[u] = h[pai] + 1;
  for (int i = 1; i < 20; i++) lca[u][i] = lca[lca[u][i - 1]][i - 1];
  parent[u] = pai;
  root[u] = root[pai];
  for (int i = 0; i < (int)lives[u].size(); i++)
    root[u] = update(root[u], 1, 101010 - 1, lives[u][i], 1);
  for (int v : g[u]) {
    if (v == pai) continue;
    dfs(v, u);
  }
}
int getLca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int d = abs(h[u] - h[v]);
  for (int i = 0; i < 20; i++)
    if (d & (1 << i)) u = lca[u][i];
  if (u == v) return u;
  for (int i = 20 - 1; i >= 0; i--) {
    if (lca[u][i] != lca[v][i]) {
      u = lca[u][i];
      v = lca[v][i];
    }
  }
  return lca[u][0];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int u, v, x;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    cin >> x;
    lives[x].push_back(i);
  }
  root[0] = 0;
  h[0] = -1;
  dfs(1, 0);
  int ans, lc, a;
  vector<int> saida;
  while (q--) {
    cin >> u >> v >> a;
    lc = getLca(u, v);
    x = parent[lc];
    saida.clear();
    for (int i = 1; i <= a; i++) {
      ans = query(root[u], root[v], root[lc], root[x], 1, 101010 - 1, i);
      if (ans == -1) break;
      saida.push_back(ans);
    }
    cout << saida.size() << " ";
    for (int i = 0; i < saida.size(); i++) cout << saida[i] << " ";
    cout << "\n";
  }
  return 0;
}