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 = 2 * 1e5 + 100, BIT = 17;
struct node {
  int a[11];
  node() { memset(a, 63, sizeof(a)); }
  void insert(int val) {
    int ind = lower_bound(a, a + 11, val) - a;
    if (ind != 11 && a[ind] == val) return;
    a[10] = val;
    sort(a, a + 11);
  }
};
int h[N];
bool vld[23];
int par[N][BIT + 10];
node par2[N][BIT + 10];
vector<int> g[N];
node a[N];
inline node mrg(const node &v1, const node &v2) {
  node ans = v1;
  for (int i = 0; i < 11; i++) ans.insert(v2.a[i]);
  return ans;
}
void dfs1(int v, int dad) {
  par[v][0] = dad;
  for (int i = 1; i <= BIT; i++)
    par[v][i] = par[v][i - 1] >= 0 ? par[par[v][i - 1]][i - 1] : -1;
  for (auto u : g[v]) {
    if (u == dad) continue;
    dfs1(u, v);
  }
}
void dfs2(int v, int dad) {
  if (dad == -1) {
    for (auto u : g[v]) dfs2(u, v);
    return;
  }
  par2[v][0] = a[dad];
  for (int i = 1; i <= BIT; i++) {
    par2[v][i] = par[v][i - 1] >= 0
                     ? mrg(par2[v][i - 1], par2[par[v][i - 1]][i - 1])
                     : node();
  }
  for (auto u : g[v])
    if (u != dad) dfs2(u, v);
}
void dfs3(int v, int dad, int he) {
  h[v] = he;
  for (auto u : g[v])
    if (u != dad) dfs3(u, v, he + 1);
}
node fnd(int v, int cnt) {
  if (cnt == 0) return node();
  int mx = -1;
  for (int i = 0; i <= BIT; i++) {
    if (cnt >= (1 << i)) mx = max(mx, i);
  }
  return mrg(par2[v][mx], fnd(par[v][mx], cnt - (1 << mx)));
}
int fnd2(int v, int cnt) {
  if (v < 0) return -1;
  if (cnt == 0) return v;
  int mx = -1;
  for (int i = 0; i <= BIT; i++)
    if (cnt >= (1 << i)) mx = max(mx, i);
  return fnd2(par[v][mx], cnt - (1 << mx));
}
int lca(int v, int u) {
  int ans = 0;
  if (h[v] > h[u]) ans += h[v] - h[u];
  if (h[v] > h[u]) swap(v, u);
  u = fnd2(u, h[u] - h[v]);
  if (v == u) return ans;
  for (int i = BIT; i >= 0; i--)
    if ((1 << i) < N && par[v][(i)] != par[u][(i)]) {
      ans += (1 << i);
      u = par[u][i];
      v = par[v][i];
    }
  ++ans;
  return ans;
}
int main() {
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    int c;
    scanf("%d", &c);
    c--;
    a[c].insert(i + 1);
  }
  dfs1(0, -1);
  dfs2(0, -1);
  dfs3(0, -1, 0);
  for (int i = 0; i < q; i++) {
    int u, v, count_;
    cin >> u >> v >> count_;
    u--;
    v--;
    int vpar = lca(v, u);
    int upar = lca(u, v);
    node ans;
    ans = fnd(v, vpar);
    if (upar > 0) {
      ans = mrg(ans, fnd(u, upar - 1));
    }
    ans = mrg(ans, a[v]);
    ans = mrg(ans, a[u]);
    int k = 0, j;
    for (j = 0; ans.a[j] <= m; j++)
      ;
    k = j;
    k = min(k, count_);
    printf("%d ", k);
    for (int i = 0; i < k; i++) printf("%d ", ans.a[i]);
    printf("\n");
  }
}