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;
struct arr {
  int a[10];
  int a_sz;
};
int n, m, q, l;
vector<vector<int> > g;
vector<vector<int> > c;
vector<int> tin, tout;
int timer;
vector<vector<int> > up;
vector<vector<arr> > mnup;
void dfs1(int v, int p = 0) {
  tin[v] = ++timer;
  up[v][0] = p;
  for (int i = 1; i <= l; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to != p) dfs1(to, v);
  }
  tout[v] = ++timer;
}
int buf[110000];
void merge(arr *a, arr *b, arr *c) {
  int *buf_end = merge(a->a, a->a + a->a_sz, b->a, b->a + b->a_sz, buf);
  buf_end = unique(buf, buf_end);
  c->a_sz = min((int)(buf_end - buf), 10);
  memcpy(c->a, buf, sizeof(int) * c->a_sz);
}
void merge2(vector<int> &a, arr *b, arr *c) {
  int *buf_end = merge(a.begin(), a.end(), b->a, b->a + b->a_sz, buf);
  buf_end = unique(buf, buf_end);
  c->a_sz = min((int)(buf_end - buf), 10);
  memcpy(c->a, buf, sizeof(int) * c->a_sz);
}
void dfs2(int v, int p = 0) {
  mnup[v][0].a_sz = 0;
  merge2(c[v], &mnup[v][0], &mnup[v][0]);
  merge2(c[p], &mnup[v][0], &mnup[v][0]);
  for (int i = 1; i <= l; ++i) {
    merge(&mnup[v][i - 1], &mnup[up[v][i - 1]][i - 1], &mnup[v][i]);
  }
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to != p) dfs2(to, v);
  }
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = l; i >= 0; --i)
    if (!upper(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
void getres(int a, int b, arr *res) {
  res->a_sz = 0;
  vector<int> ca(c[a].begin(), c[a].begin() + min((int)c[a].size(), 10));
  vector<int> cb(c[b].begin(), c[b].begin() + min((int)c[b].size(), 10));
  merge2(ca, res, res);
  merge2(cb, res, res);
  for (int i = l; i >= 0; --i) {
    if (!upper(up[a][i], b)) {
      merge(res, &mnup[a][i], res);
      a = up[a][i];
    }
  }
}
int main() {
  cin >> n >> m >> q;
  g.resize(n);
  c.resize(n);
  tin.resize(n), tout.resize(n), up.resize(n), mnup.resize(n);
  for (int(i) = (0); (i) < (int)(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) < (int)(m); ++(i)) {
    int v;
    cin >> v;
    c[v - 1].push_back(i + 1);
  }
  for (int(i) = (0); (i) < (int)(n); ++(i)) {
    sort(c[i].begin(), c[i].end());
  }
  l = 1;
  while ((1 << l) <= n) ++l;
  for (int i = 0; i < n; ++i) {
    up[i].resize(l + 1);
    mnup[i].resize(l + 1);
  }
  dfs1(0);
  dfs2(0);
  for (int(i) = (0); (i) < (int)(q); ++(i)) {
    int a, b, aa;
    cin >> a >> b >> aa;
    a--, b--;
    arr res, res2;
    if (upper(a, b)) {
      getres(b, a, &res);
    } else if (upper(b, a)) {
      getres(a, b, &res);
    } else {
      int p = lca(a, b);
      getres(a, p, &res);
      getres(b, p, &res2);
      merge(&res, &res2, &res);
    }
    int k = min(res.a_sz, aa);
    printf("%d ", k);
    for (int(j) = (0); (j) < (int)(k); ++(j)) {
      printf("%d ", res.a[j]);
    }
    printf("\n");
  }
  return 0;
}