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 maxk = 10;
const int maxn = 100100;
vector<int> adj[maxn];
int par[20][maxn];
int depth[maxn];
void dfs(int a, int p = 0) {
  par[0][a] = p;
  for (int i = 0; i < adj[a].size(); i++) {
    int b = adj[a][i];
    if (b == p) continue;
    depth[b] = depth[a] + 1;
    dfs(b, a);
  }
}
void merge(const vector<int>& va, const vector<int>& vb, vector<int>& res,
           int k) {
  int a = 0, b = 0;
  while (res.size() < k) {
    if (a < va.size() && (b >= vb.size() || va[a] < vb[b])) {
      res.push_back(va[a]);
      a++;
    } else if (b < vb.size()) {
      res.push_back(vb[b]);
      b++;
    } else {
      break;
    }
  }
}
vector<int> p[20][maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(0);
  for (int i = 0; i < m; i++) {
    int a;
    cin >> a;
    a--;
    p[0][a].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    sort(p[0][i].begin(), p[0][i].end());
    if (maxk < p[0][i].size()) {
      p[0][i].resize(maxk);
    }
  }
  for (int i = 0; i < 19; i++) {
    for (int j = 0; j < n; j++) {
      par[i + 1][j] = par[i][par[i][j]];
      merge(p[i][j], p[i][par[i][j]], p[i + 1][j], maxk);
    }
  }
  for (int i = 0; i < q; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    int da = depth[a];
    int db = depth[b];
    if (da < db) {
      swap(a, b);
      swap(da, db);
    }
    int diff = da - db;
    vector<int> res;
    for (int i = 0; i < 20; i++) {
      if (diff & (1 << i)) {
        vector<int> t;
        merge(res, p[i][a], t, c);
        swap(res, t);
        a = par[i][a];
      }
    }
    if (a != b) {
      for (int i = 19; i >= 0; i--) {
        if (par[i][a] != par[i][b]) {
          {
            vector<int> t;
            merge(res, p[i][a], t, c);
            swap(res, t);
          }
          {
            vector<int> t;
            merge(res, p[i][b], t, c);
            swap(res, t);
          }
          a = par[i][a];
          b = par[i][b];
        }
      }
      {
        vector<int> t;
        merge(res, p[0][a], t, c);
        swap(res, t);
        a = par[0][a];
      }
      {
        vector<int> t;
        merge(res, p[0][b], t, c);
        swap(res, t);
        b = par[0][b];
      }
    }
    {
      vector<int> t;
      merge(res, p[0][a], t, c);
      swap(res, t);
    }
    cout << res.size();
    for (int i = 0; i < res.size(); i++) {
      cout << ' ' << res[i] + 1;
    }
    cout << '\n';
  }
}