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>
const int MOD = int(1e9) + 7;
inline int add(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline void inc(int& a, int b) { a = add(a, b); }
inline int sub(int a, int b) { return (a - b < 0) ? (a - b + MOD) : (a - b); }
inline void dec(int& a, int b) { a = sub(a, b); }
inline int mul(int a, int b) { return (a * 1ll * b) % MOD; }
using namespace std;
const int N = 100000, LOG = 20, MAX = 10;
vector<int> ladj[N], V[N][LOG], people[N];
int P[N][LOG], H[N];
vector<int> join(vector<int>& a, vector<int>& b, int limit) {
  vector<int> ret;
  int i = 0, j = 0;
  while (i < ((int)(a).size()) && j < ((int)(b).size()) &&
         ((int)(ret).size()) < limit) {
    if (a[i] == b[j])
      ret.push_back(a[i++, j++]);
    else if (a[i] < b[j])
      ret.push_back(a[i++]);
    else
      ret.push_back(b[j++]);
  }
  while (i < ((int)(a).size()) && ((int)(ret).size()) < limit)
    ret.push_back(a[i++]);
  while (j < ((int)(b).size()) && ((int)(ret).size()) < limit)
    ret.push_back(b[j++]);
  return ret;
}
void dfs(int u, int p) {
  P[u][0] = p;
  for (__typeof((ladj[u]).begin()) it = (ladj[u]).begin();
       it != (ladj[u]).end(); ++it) {
    int v = *it;
    if (v != p) {
      H[v] = H[u] + 1;
      dfs(v, u);
    }
  }
  sort(people[u].begin(), people[u].end());
  for (int i = 0; i < ((int)(people[u]).size()) && i < MAX; i++)
    V[u][0].push_back(people[u][i]);
  people[u].clear();
}
vector<int> lca(int u, int v, int a) {
  if (H[u] < H[v]) swap(u, v);
  vector<int> ret;
  for (int i = LOG - 1; i >= 0; i--)
    if (P[u][i] != -1 && H[P[u][i]] >= H[v]) {
      ret = join(ret, V[u][i], a);
      u = P[u][i];
    }
  if (u == v) {
    ret = join(ret, V[u][0], a);
    return ret;
  }
  for (int i = LOG - 1; i >= 0; i--)
    if (P[u][i] != P[v][i]) {
      ret = join(ret, V[u][i], a);
      ret = join(ret, V[v][i], a);
      u = P[u][i];
      v = P[v][i];
    }
  ret = join(ret, V[u][0], a);
  ret = join(ret, V[v][0], a);
  u = P[u][0];
  ret = join(ret, V[u][0], a);
  return ret;
}
bool solve(int testNumber) {
  int n, m, q;
  if (scanf("%d %d %d", &n, &m, &q) == EOF) return false;
  for (int i = 0; i < N; i++) {
    ladj[i].clear();
    for (int j = 0; j < LOG; j++) V[i][j].clear();
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    ladj[u].push_back(v);
    ladj[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    people[x - 1].push_back(i + 1);
  }
  H[0] = 0;
  dfs(0, -1);
  for (int j = 1; j < LOG; j++)
    for (int i = 0; i < n; i++) {
      if (P[i][j - 1] == -1) {
        P[i][j] = -1;
        continue;
      }
      P[i][j] = P[P[i][j - 1]][j - 1];
      if (P[i][j] != -1)
        V[i][j] = join(V[i][j - 1], V[P[i][j - 1]][j - 1], MAX);
    }
  while (q-- > 0) {
    int u, v, a;
    scanf("%d %d %d", &u, &v, &a), u--, v--;
    vector<int> aux = lca(u, v, a);
    printf("%d", ((int)(aux).size()));
    for (__typeof((aux).begin()) it = (aux).begin(); it != (aux).end(); ++it)
      printf(" %d", *it);
    puts("");
  }
  return true;
}
int main() {
  for (int i = 1; solve(i); i++)
    ;
  return 0;
}