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 MAXN = 1e5;
const int MAXLOG = 20;
vector<int> merge(vector<int> a, vector<int> b) {
  if (a.size() == 0) return b;
  if (b.size() == 0) return a;
  vector<int> ret(a.size() + b.size());
  merge(a.begin(), a.end(), b.begin(), b.end(), ret.begin());
  vector<int>::iterator it = unique(ret.begin(), ret.end());
  ret.resize(min((int)distance(ret.begin(), it), 10));
  return ret;
}
int N;
vector<int> G[MAXN], p[MAXN][MAXLOG];
int H[MAXN], anc[MAXN][MAXLOG];
void pre_dfs(int u, int P = -1) {
  anc[u][0] = P;
  if (P != -1)
    H[u] = H[P] + 1;
  else
    H[u] = 0;
  for (int i = 1; i < MAXLOG; i++)
    if (anc[u][i - 1] != -1) {
      anc[u][i] = anc[anc[u][i - 1]][i - 1];
      p[u][i] = merge(p[u][i - 1], p[anc[u][i - 1]][i - 1]);
    }
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (P != v) {
      pre_dfs(v, u);
    }
  }
}
vector<int> LCA(int u, int v) {
  vector<int> ret;
  if (H[v] < H[u]) swap(v, u);
  for (int i = MAXLOG - 1; i > -1; i--)
    if (anc[v][i] != -1 && H[anc[v][i]] >= H[u]) {
      ret = merge(ret, p[v][i]);
      v = anc[v][i];
    }
  if (v == u) {
    ret = merge(ret, p[v][0]);
    return ret;
  }
  for (int i = MAXLOG - 1; i > -1; i--)
    if (anc[v][i] != anc[u][i]) {
      ret = merge(ret, merge(p[v][i], p[u][i]));
      v = anc[v][i], u = anc[u][i];
    }
  return merge(ret, merge(p[v][1], p[u][1]));
}
int main() {
  int M, Q;
  scanf("%d %d", &N, &M);
  scanf("%d", &Q);
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 0; i < M; i++) {
    int u;
    scanf("%d", &u);
    u--;
    p[u][0].push_back(i);
  }
  for (int i = 0; i < N; i++) sort(p[i][0].begin(), p[i][0].end());
  memset(anc, -1, sizeof(anc));
  pre_dfs(0);
  for (int i = 0; i < Q; i++) {
    int u, v, a;
    scanf("%d %d", &u, &v);
    scanf("%d", &a);
    u--;
    v--;
    vector<int> ret = LCA(u, v);
    printf("%d ", min((int)ret.size(), a));
    for (int i = 0; i < min((int)ret.size(), a); i++) {
      printf("%d ", ret[i] + 1);
    }
    printf("\n");
  }
}