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;
int anc[100004][40], L[100004];
int path[100004][40][12], *buf = new int[12];
vector<int> adj[100004];
int n, m, q, u, v, a;
void merge(int *a, int *b) {
  int k = 0, i = 0, j = 0;
  for (; a[i] != -1 && b[j] != -1 && k < 10;) {
    if (a[i] == b[j])
      buf[k++] = a[i], i++, j++;
    else if (a[i] < b[j])
      buf[k++] = a[i++];
    else
      buf[k++] = b[j++];
  }
  while (k < 10 && a[i] != -1) buf[k++] = a[i++];
  while (k < 10 && b[j] != -1) buf[k++] = b[j++];
  for (int i = 0; i < k; ++i) a[i] = buf[i];
  a[k] = -1;
}
void dfs(int cur, int par) {
  anc[cur][0] = cur;
  anc[cur][1] = par;
  L[cur] = L[par] + 1;
  for (int ch : adj[cur])
    if (ch != par) dfs(ch, cur);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  memset(path, -1, sizeof path);
  for (int i = 1; i < n; ++i)
    scanf("%d%d", &u, &v), adj[u].push_back(v), adj[v].push_back(u);
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &u);
    int sz = 0;
    while (path[u][0][sz] != -1) sz++;
    if (sz < 10) path[u][0][sz] = i;
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i)
    merge(path[i][1], path[i][0]), merge(path[i][1], path[anc[i][1]][0]);
  for (int lv = 2; lv < 40; ++lv) {
    for (int i = 1; i <= n; ++i) {
      int prv = anc[i][lv - 1];
      anc[i][lv] = anc[prv][lv - 1];
      merge(path[i][lv], path[i][lv - 1]);
      merge(path[i][lv], path[prv][lv - 1]);
    }
  }
  int x[12], y[12];
  while (q--) {
    scanf("%d%d%d", &u, &v, &a), x[0] = y[0] = -1;
    if (L[u] > L[v]) swap(u, v);
    for (int i = 1, d = L[v] - L[u]; d; ++i)
      if (d & (1 << (i - 1))) {
        d ^= (1 << (i - 1)), merge(y, path[v][i]), v = anc[v][i];
      }
    for (int up = 20; up && u != v; up--) {
      if (anc[u][up] != anc[v][up])
        merge(x, path[u][up]), u = anc[u][up], merge(y, path[v][up]),
                               v = anc[v][up];
    }
    if (u != v)
      merge(x, path[u][1]), u = anc[u][1], merge(y, path[v][1]), v = anc[v][1];
    merge(x, y);
    merge(x, path[u][0]);
    int sz = 0;
    while (x[sz] != -1) sz++;
    if (a > sz) a = sz;
    printf("%d", a);
    for (int i = 0; i < a; ++i) printf(" %d", x[i]);
    puts("");
  }
}
