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 MAX_V = 100010;
const int MAX_LOG_V = 20;
int n, m;
vector<int> G[MAX_V];
vector<int> persons[MAX_V];
int parent[MAX_LOG_V][MAX_V];
vector<int> seq[MAX_LOG_V][MAX_V];
int depth[MAX_V];
inline void merge(vector<int>& a, const vector<int>& b) {
  vector<int> tmp(10);
  int i = 0, j = 0, k = 0;
  ;
  while (i < a.size() && j < b.size() && k < tmp.size()) {
    if (a[i] < b[j]) {
      tmp[k++] = a[i++];
    } else {
      tmp[k++] = b[j++];
    }
  }
  while (k < tmp.size() && i < a.size()) {
    tmp[k++] = a[i++];
  }
  while (k < tmp.size() && j < b.size()) {
    tmp[k++] = b[j++];
  }
  tmp.resize(k);
  a.swap(tmp);
}
void dfs(int v, int p, int d) {
  parent[0][v] = p;
  merge(seq[0][v], persons[v]);
  depth[v] = d;
  for (int u : G[v]) {
    if (u != p) {
      dfs(u, v, d + 1);
    }
  }
}
void init() {
  dfs(0, -1, 0);
  for (int k = 0; k + 1 < MAX_LOG_V; k++) {
    for (int v = 0; v < n; v++) {
      if (parent[k][v] < 0) {
        parent[k + 1][v] = -1;
        merge(seq[k + 1][v], seq[k][v]);
      } else {
        int p = parent[k][v];
        parent[k + 1][v] = parent[k][p];
        merge(seq[k + 1][v], seq[k][v]);
        merge(seq[k + 1][v], seq[k][p]);
      }
    }
  }
}
vector<int> solve(int u, int v, int a) {
  vector<int> res;
  if (depth[u] > depth[v]) swap(u, v);
  for (int k = 0; k < MAX_LOG_V; k++) {
    if ((depth[v] - depth[u]) >> k & 1) {
      merge(res, seq[k][v]);
      v = parent[k][v];
    }
  }
  if (u != v) {
    for (int k = MAX_LOG_V - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        merge(res, seq[k][u]);
        merge(res, seq[k][v]);
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    int lca = parent[0][u];
    merge(res, seq[0][u]);
    merge(res, seq[0][v]);
    merge(res, seq[0][lca]);
  } else {
    merge(res, seq[0][u]);
  }
  int sz = min((int)res.size(), a);
  res.resize(sz);
  return res;
}
int main() {
  int q;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    scanf("%d %d", &v, &u);
    --v;
    --u;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    scanf("%d", &c);
    --c;
    if (persons[c].size() < 10) {
      persons[c].push_back(i);
    }
  }
  init();
  for (int i = 0; i < q; i++) {
    int u, v, a;
    scanf("%d %d %d", &u, &v, &a);
    --u;
    --v;
    vector<int> res = solve(u, v, a);
    printf("%d", (int)res.size());
    for (int x : res) {
      printf(" %d", x);
    }
    puts("");
  }
  return 0;
}