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>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100, maxl = 20, vsize = 10;
struct vals {
  int n = 0;
  int a[10] = {0};
  inline void pop_back() {
    if (n) --n;
  }
  inline int &operator[](const int &x) { return a[x]; }
  inline int back() {
    if (n) return a[n - 1];
    return -1;
  }
  inline void push_back(const int &x) {
    if (n < 10 && x != back()) a[n++] = x;
  }
  inline int size() { return n; }
  inline bool empty() { return !n; }
} vec[maxn][maxl], ins[maxn];
vector<int> adj[maxn];
int par[maxn][maxl], h[maxn];
inline vals operator+(vals &v, vals &u) {
  vals ans;
  int i = 0, j = 0;
  while (i < v.size() && j < u.size()) {
    if (v[i] <= u[j])
      ans.push_back(v[i++]);
    else
      ans.push_back(u[j++]);
  }
  while (i < v.size()) ans.push_back(v[i++]);
  while (j < u.size()) ans.push_back(u[j++]);
  return ans;
}
inline void dfs(int v = 0, int p = -1) {
  par[v][0] = p;
  vec[v][0] = ins[v];
  if (p + 1) {
    h[v] = h[p] + 1;
    vec[v][0] = ins[v] + ins[p];
  }
  for (int i = 1; i < maxl; ++i) {
    if (par[v][i - 1] + 1) {
      par[v][i] = par[par[v][i - 1]][i - 1];
      vec[v][i] = vec[v][i - 1] + vec[par[v][i - 1]][i - 1];
    }
  }
  for (auto &u : adj[v])
    if (p - u) dfs(u, v);
}
inline vals lca(int v, int u) {
  vals ans = ins[v] + ins[u];
  if (h[v] < h[u]) swap(v, u);
  for (int i = maxl - 1; i > -1; --i) {
    if (par[v][i] + 1 && h[par[v][i]] >= h[u]) {
      ans = ans + vec[v][i];
      v = par[v][i];
    }
  }
  if (v == u) return ans;
  for (int i = maxl - 1; i > -1; --i) {
    if (par[v][i] - par[u][i]) {
      ans = ans + vec[v][i];
      ans = ans + vec[u][i];
      v = par[v][i], u = par[u][i];
    }
  }
  ans = ans + ins[par[v][0]];
  return ans;
}
int n, m, q;
int v, u, a;
int main() {
  memset(par, -1, sizeof par);
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    int v, u;
    scanf("%d %d", &v, &u);
    --v, --u;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  for (int i = 1; i < m + 1; ++i) {
    int v;
    scanf("%d", &v);
    --v;
    if ((int)ins[v].size() < vsize) ins[v].push_back(i);
  }
  dfs();
  while (q--) {
    scanf("%d%d%d", &v, &u, &a);
    --v, --u;
    vals ans = lca(v, u);
    if (ans.size() > a) ans.n = a;
    vector<int> vec;
    vec.push_back(ans.size());
    for (int i = 0; i < ans.size(); ++i) vec.push_back(ans[i]);
    for (int i = 0; i < vec.size(); ++i) {
      printf("%d", vec[i]);
      if (i + 1 == vec.size())
        puts("");
      else
        printf(" ");
    }
  }
  return 0;
}
