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;
const int nax = 100005, mod = 1e9 + 7;
const long long inf = 1e18;
int n, m, q;
int a, b;
vector<int> v[nax];
vector<int> d[nax];
int h[nax], lca[nax][17];
vector<int> s[nax][17];
void dfs(int node, int par) {
  for (auto it : v[node]) {
    if (it == par) continue;
    h[it] = h[node] + 1;
    lca[it][0] = node;
    dfs(it, node);
  }
  for (auto it : d[node]) {
    s[node][0].push_back(it);
  }
  sort(s[node][0].begin(), s[node][0].end());
  while ((int)s[node][0].size() > 10) {
    s[node][0].pop_back();
  }
}
int daj(int x, int y) {
  if (h[x] < h[y]) swap(x, y);
  for (int i = 16; 0 <= i; --i) {
    if (h[x] - (1 << i) >= h[y]) x = lca[x][i];
  }
  if (x == y) return x;
  for (int i = 16; 0 <= i; --i) {
    if (lca[x][i] != lca[y][i]) {
      x = lca[x][i];
      y = lca[y][i];
    }
  }
  return lca[x][0];
}
void add(vector<int> &C, vector<int> &A, vector<int> &B) {
  for (auto it : A) C.push_back(it);
  for (auto it : B) C.push_back(it);
  sort(C.begin(), C.end());
  while ((int)C.size() > 10) {
    C.pop_back();
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; n - 1 >= i; ++i) {
    scanf("%d %d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &a);
    d[a].push_back(i);
  }
  dfs(1, -1);
  for (int j = 1; 16 >= j; ++j)
    for (int i = 1; n >= i; ++i) {
      lca[i][j] = lca[lca[i][j - 1]][j - 1];
      add(s[i][j], s[i][j - 1], s[lca[i][j - 1]][j - 1]);
    }
  while (q--) {
    int ile;
    scanf("%d %d %d", &a, &b, &ile);
    int e = daj(a, b);
    int ha = h[a] - h[e];
    int hb = h[b] - h[e];
    int aa, bb;
    aa = a;
    bb = b;
    vector<int> w;
    for (int i = 0; 16 >= i; ++i) {
      if (ha & (1 << i)) {
        for (auto it : s[aa][i]) w.push_back(it);
        aa = lca[aa][i];
      }
      if (hb & (1 << i)) {
        for (auto it : s[bb][i]) w.push_back(it);
        bb = lca[bb][i];
      }
    }
    for (auto it : s[e][0]) w.push_back(it);
    sort(w.begin(), w.end());
    printf("%d ", min((int)w.size(), ile));
    while ((int)w.size() > 10 || (int)w.size() > ile) w.pop_back();
    for (int i = 0; min((int)w.size(), ile) > i; ++i) {
      printf("%d ", w[i]);
    }
    printf("\n");
  }
  return 0;
}
