587_C. Duff in the Army

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>

Example 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


Example output:
1 3 
2 2 3 
0 
3 1 2 4 
1 2 


Tags: data structures
trees
Difficulty rating: 2200

Solution:
#include <bits/stdc++.h>
const int N = 1e5 + 1;
int i, j, n, m, q, t, d[N], lg[N], fa[N], en[N * 2], nxt[N * 2], ft[N][17];
std::priority_queue<int> b, s, f[N][17];
void Mer(std::priority_queue<int>& a, std::priority_queue<int>& x) {
  for (b = x; !b.empty(); a.size() > t ? a.pop() : void(0))
    a.push(b.top()), b.pop();
}
void dfs(int u) {
  for (i = 0; i++ < lg[d[u]];
       Mer(f[u][i] = f[u][i - 1], f[ft[u][i - 1]][i - 1]))
    ft[u][i] = ft[ft[u][i - 1]][i - 1];
  for (int i = fa[u]; i; i = nxt[i])
    if (!d[en[i]]) d[en[i]] = d[u] + 1, *ft[en[i]] = u, dfs(en[i]);
}
void cal(int x, int y) {
  if (d[x] < d[y]) std::swap(x, y);
  for (s = **f, n = d[x] - d[y]; n; n -= n & -n)
    Mer(s, f[x][lg[n & -n]]), x = ft[x][lg[n & -n]];
  if (x == y) return Mer(s, *f[x]);
  for (n = lg[d[x]]; ~n; --n)
    if (ft[x][n] ^ ft[y][n])
      Mer(s, f[x][n]), x = ft[x][n], Mer(s, f[y][n]), y = ft[y][n];
  Mer(s, f[x][1]), Mer(s, *f[y]);
}
void W(int n, int x) {
  if (!n) return;
  s.pop(), W(n - 1, s.top());
  if (n <= t) printf(" %d", x);
}
int main() {
  for (scanf("%d%d%d", &n, &m, &q); ++t < n + n - 1;) {
    scanf("%d%d", &i, &j);
    en[t] = j, nxt[t] = fa[i], fa[i] = t;
    en[++t] = i, nxt[t] = fa[j], fa[j] = t;
  }
  for (i = 1; i++ < n; lg[i] = lg[i >> 1] + 1)
    ;
  for (t = 10, i = 0; i++ < m;)
    if (scanf("%d", &j), f[j]->push(i), f[j]->size() > t) f[j]->pop();
  for (dfs(d[1] = 1); q--; puts("")) {
    scanf("%d%d%d", &i, &j, &t), cal(i, j);
    printf("%d", t = std::min((int)s.size(), t));
    W(s.size(), s.top());
  }
}