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;
const int N = 100010;
struct node {
  int d[12];
  void init() {
    for (int i = 0; i < 12; ++i) d[i] = 0;
  }
  void init(int x) {
    init();
    d[0] = 1;
    d[1] = x;
  }
  void reset() {
    while (d[0] > 0 && d[d[0]] == 0) d[0]--;
  }
};
int n, m, query, p[N][20], q[N], L[N];
vector<int> ke[N];
node a[N], A[N][20], ans;
node combine(node &a, node &b) {
  node c;
  c.init();
  int i = 1, j = 1;
  while (c.d[0] < 10 && i <= a.d[0] && j <= b.d[0]) {
    c.d[0]++;
    if (a.d[i] < b.d[j])
      c.d[c.d[0]] = a.d[i], i++;
    else
      c.d[c.d[0]] = b.d[j], j++;
  }
  while (c.d[0] < 10 && i <= a.d[0]) c.d[0]++, c.d[c.d[0]] = a.d[i], i++;
  while (c.d[0] < 10 && j <= b.d[0]) c.d[0]++, c.d[c.d[0]] = b.d[j], j++;
  c.reset();
  return c;
}
void build_tree() {
  int l = 1, r = 1, u, v;
  q[1] = 1;
  p[1][0] = -1;
  L[1] = 0;
  while (l <= r) {
    u = q[l];
    l++;
    for (vector<int>::iterator i = ke[u].begin(); i != ke[u].end(); ++i)
      if (*i != p[u][0]) {
        p[*i][0] = u;
        L[*i] = L[u] + 1;
        r++;
        q[r] = *i;
      }
  }
}
void enter() {
  scanf("%d%d%d", &n, &m, &query);
  int u, v;
  for (int _ = 0; _ < n - 1; ++_) {
    scanf("%d%d", &u, &v);
    ke[u].push_back(v);
    ke[v].push_back(u);
  }
  memset(p, 255, sizeof(p));
  build_tree();
  for (int _ = 0; _ < m; ++_) {
    scanf("%d", &u);
    node tmp;
    tmp.init(_ + 1);
    a[u] = combine(a[u], tmp);
  }
}
void pre_LCA() {
  for (int i = 1; i <= n; ++i) A[i][0] = a[i];
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = 1; i <= n; ++i)
      if (p[i][j - 1] != -1)
        p[i][j] = p[p[i][j - 1]][j - 1],
        A[i][j] = combine(A[i][j - 1], A[p[i][j - 1]][j - 1]);
}
void LCA(int u, int v) {
  ans.init();
  if (L[u] < L[v]) swap(u, v);
  int k = 0;
  for (k = 1; (1 << k) <= L[u]; ++k)
    ;
  k--;
  for (int i = k; i >= 0; --i)
    if (L[u] - (1 << i) >= L[v]) ans = combine(ans, A[u][i]), u = p[u][i];
  if (u == v) {
    ans = combine(ans, A[u][0]);
    return;
  }
  for (int i = k; i >= 0; --i)
    if (p[u][i] != -1 && p[u][i] != p[v][i])
      ans = combine(ans, A[u][i]), u = p[u][i], ans = combine(ans, A[v][i]),
      v = p[v][i];
  ans = combine(ans, A[u][0]);
  ans = combine(ans, A[v][0]);
  ans = combine(ans, a[p[u][0]]);
}
void process() {
  pre_LCA();
  int u, v, k;
  for (int _ = 0; _ < query; ++_) {
    scanf("%d%d%d", &u, &v, &k);
    LCA(u, v);
    k = min(k, ans.d[0]);
    printf("%d%c", k, (k == 0) ? '\n' : ' ');
    for (int i = 1; i <= k; ++i)
      printf("%d%c", ans.d[i], (i == k) ? '\n' : ' ');
  }
}
int main() {
  enter();
  process();
  return 0;
}