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 = 100000;
const int LOGMAX = 20;
int p[MAX][LOGMAX];
int h[MAX];
vector<int> s[MAX][LOGMAX];
vector<int> g[MAX];
vector<int> merge(const vector<int> &a, const vector<int> &b) {
  vector<int> r;
  int i = 0, j = 0;
  while (r.size() < 10 && (i < a.size() || j < b.size())) {
    if (i == a.size()) {
      r.push_back(b[j]);
      j++;
    } else if (j == b.size() || a[i] < b[j]) {
      r.push_back(a[i]);
      i++;
    } else {
      r.push_back(b[j]);
      j++;
    }
  }
  return r;
}
void dfs(int u, int par, int depth) {
  h[u] = depth;
  p[u][0] = par;
  for (int i = 1; p[u][i - 1] != -1; i++) {
    p[u][i] = p[p[u][i - 1]][i - 1];
    s[u][i] = merge(s[u][i - 1], s[p[u][i - 1]][i - 1]);
  }
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v != par) {
      dfs(g[u][i], u, depth + 1);
    }
  }
}
vector<int> query(int a, int b) {
  if (h[a] < h[b]) {
    swap(a, b);
  }
  vector<int> r;
  for (int i = LOGMAX - 1; h[b] < h[a]; i--) {
    if (p[a][i] != -1 && h[p[a][i]] >= h[b]) {
      r = merge(r, s[a][i]);
      a = p[a][i];
    }
  }
  int f = 0;
  for (int i = LOGMAX - 1; p[a][0] != p[b][0]; i--) {
    if (p[a][i] != p[b][i]) {
      r = merge(r, merge(s[a][i], s[b][i]));
      a = p[a][i];
      b = p[b][i];
    }
  }
  if (a == b) {
    r = merge(r, s[a][0]);
  } else {
    r = merge(r, merge(s[a][1], s[b][0]));
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  memset(p, -1, MAX * LOGMAX * sizeof(int));
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 0; i < m; i++) {
    int c;
    cin >> c;
    s[c - 1][0].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) {
    sort(s[i][0].begin(), s[i][0].end());
  }
  dfs(0, -1, 0);
  for (int i = 0; i < q; i++) {
    int v, u, a;
    cin >> v >> u >> a;
    v--;
    u--;
    vector<int> r = query(v, u);
    int k = min(a, (int)r.size());
    cout << k << ' ';
    for (int j = 0; j < k; j++) {
      cout << r[j] << ' ';
    }
    cout << endl;
  }
  return 0;
}