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 N = 1e5 + 100;
const int M = 1e5;
const long long mod = 998244353;
const int inf = 2e9 + 3;
const long long INF = 1e18;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
void data() {}
int n, m, q, up[N][100], tin[N], tout[N], tick;
vector<int> g[N], num[N];
struct tree {
  int sum;
  tree *l, *r;
  tree(int new_sum = 0) {
    l = r = nullptr;
    sum = new_sum;
  }
  tree(tree *tl, tree *tr) {
    l = tl;
    r = tr;
    sum = 0;
    if (l) sum += l->sum;
    if (r) sum += r->sum;
  }
  tree(tree *v) {
    sum = v->sum;
    l = v->l;
    r = v->r;
  }
} * t[N];
tree *build(int tl = 1, int tr = m) {
  if (tl == tr) return new tree(0);
  int tm = (tl + tr) / 2;
  return new tree(build(tl, tm), build(tm + 1, tr));
}
tree *upd(int pos, tree *v, int tl = 1, int tr = m) {
  if (tl == tr) {
    return new tree(1);
  }
  int tm = (tl + tr) / 2;
  tree *cnt = new tree(v);
  if (pos <= tm) {
    cnt->l = upd(pos, v->l, tl, tm);
  } else {
    cnt->r = upd(pos, v->r, tm + 1, tr);
  }
  cnt->sum = cnt->l->sum + cnt->r->sum;
  return cnt;
}
int get(int k, tree *a, tree *b, tree *v, tree *pv, int tl = 1, int tr = m) {
  if (tl == tr) return tl;
  int tm = (tl + tr) / 2;
  int left = a->l->sum + b->l->sum - v->l->sum - pv->l->sum;
  if (left < k) {
    return get(k - left, a->r, b->r, v->r, pv->r, tm + 1, tr);
  } else {
    return get(k, a->l, b->l, v->l, pv->l, tl, tm);
  }
}
void dfs(int v = 1, int p = 0) {
  up[v][0] = p;
  for (int i = 1; i <= 20; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  tin[v] = ++tick;
  t[v] = t[p];
  for (auto it : num[v]) t[v] = upd(it, t[v]);
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
  tout[v] = ++tick;
}
bool upper(int v, int u) { return tin[v] <= tin[u] && tout[u] <= tout[v]; }
int lca(int v, int u) {
  if (upper(v, u)) return v;
  if (upper(u, v)) return u;
  for (int i = 20; i >= 0; --i) {
    if (!upper(up[v][i], u) && up[v][i]) v = up[v][i];
  }
  return up[v][0];
}
int main() {
  data();
  scanf("%d %d %d", &n, &m, &q);
  t[0] = build();
  for (int i = 1, v, u; i < n; ++i) {
    scanf("%d %d", &v, &u);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1, c; i <= m; ++i) {
    scanf("%d", &c);
    num[c].push_back(i);
  }
  dfs();
  while (q--) {
    int v, u, a;
    scanf("%d %d %d", &v, &u, &a);
    vector<int> ans;
    int lc = lca(v, u);
    for (int i = 1;
         i <= min(a, t[v]->sum + t[u]->sum - t[lc]->sum - t[up[lc][0]]->sum);
         ++i) {
      int nw = get(i, t[v], t[u], t[lc], t[up[lc][0]]);
      if ((int)ans.size() && nw == ans.back()) break;
      ans.push_back(nw);
    }
    printf("%d ", (int)ans.size());
    for (auto it : ans) printf("%d ", it);
    printf("\n");
  }
}