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;
vector<int> g[100006];
vector<int> located[100006];
int parent[100006], level[100006], visited[100006];
int ancestor[100006][18];
vector<int> st[100006][18];
void dfs(int u) {
  visited[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (visited[v] == 0) {
      parent[v] = u;
      level[v] = level[u] + 1;
      dfs(v);
    }
  }
}
void merge(vector<int>& a, vector<int>& b) {
  vector<int> tmp;
  int cnt = 0;
  int l1 = a.size(), l2 = b.size();
  int i = 0, j = 0;
  while (i < l1 && j < l2 && cnt < 10) {
    if (a[i] < b[j])
      tmp.push_back(a[i++]);
    else
      tmp.push_back(b[j++]);
    cnt++;
  }
  while (i < l1 && cnt < 10) {
    tmp.push_back(a[i++]);
    cnt++;
  }
  while (j < l2 && cnt < 10) {
    tmp.push_back(b[j++]);
    cnt++;
  }
  a.clear();
  for (i = 0; i < cnt; i++) a.push_back(tmp[i]);
}
int main() {
  int tc = 1;
  for (int t = 1; t <= tc; t++) {
    int n, m, q;
    scanf("%d", &(n));
    scanf("%d", &(m));
    scanf("%d", &(q));
    for (int j = 1; j < 18; j++) {
      for (int i = 1; i <= n; i++) {
        ancestor[i][j] = -1;
      }
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d", &(u));
      scanf("%d", &(v));
      g[u].push_back(v);
      g[v].push_back(u);
    }
    for (int i = 1; i <= m; i++) {
      int tmp;
      scanf("%d", &(tmp));
      located[tmp].push_back(i);
    }
    for (int i = 1; i <= n; i++) sort(located[i].begin(), located[i].end());
    parent[1] = -1;
    level[1] = 1;
    dfs(1);
    for (int i = 1; i <= n; i++) ancestor[i][0] = parent[i];
    for (int j = 1; j < 18; j++) {
      for (int i = 1; i <= n; i++) {
        if (ancestor[i][j - 1] != -1)
          ancestor[i][j] = ancestor[ancestor[i][j - 1]][j - 1];
      }
    }
    for (int i = 1; i <= n; i++) {
      st[i][0] = located[i];
      vector<int> vi;
      merge(st[i][0], vi);
    }
    for (int j = 1; j < 18; j++) {
      for (int i = 1; i <= n; i++) {
        st[i][j] = st[i][j - 1];
        if (ancestor[i][j - 1] == -1) continue;
        merge(st[i][j], st[ancestor[i][j - 1]][j - 1]);
      }
    }
    while (q--) {
      int u, v, a;
      scanf("%d", &(u));
      scanf("%d", &(v));
      scanf("%d", &(a));
      if (level[u] < level[v]) {
        int tmp = u;
        u = v;
        v = tmp;
      }
      vector<int> tmp;
      for (int j = 17; j >= 0; j--) {
        if (ancestor[u][j] != -1 && level[ancestor[u][j]] >= level[v]) {
          merge(tmp, st[u][j]);
          u = ancestor[u][j];
        }
      }
      if (u != v) {
        for (int j = 17; j >= 0; j--) {
          if (ancestor[u][j] != -1 && ancestor[u][j] != ancestor[v][j]) {
            merge(tmp, st[u][j]);
            merge(tmp, st[v][j]);
            u = ancestor[u][j];
            v = ancestor[v][j];
          }
        }
        merge(tmp, st[u][0]);
        merge(tmp, st[v][0]);
        u = parent[u];
        v = parent[v];
      }
      merge(tmp, st[u][0]);
      int cnt = 0;
      int len = tmp.size();
      cout << min(a, len);
      for (int it = 0; it < tmp.size() && cnt < a; it++, cnt++) {
        printf(" ");
        printf("%d", (tmp[it]));
      }
      printf("\n");
    }
  }
}