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 INF = 1000000002;
const long long INFLL = 100000000000000000ll;
const long long MOD = 1000000007;
int D[100005], L[100005][17], N, M, Q;
vector<int> P[100005][17];
vector<int> E[100005];
void dfs(int n, int p, int d) {
  D[n] = d;
  L[n][0] = p;
  for (int v : E[n])
    if (v != p) dfs(v, n, d + 1);
}
void merge(vector<int> &v, vector<int> &aux) {
  for (int k : aux) v.push_back(k);
  sort(v.begin(), v.end());
  while (((int)v.size()) > 10) v.pop_back();
}
vector<int> lca(int u, int v) {
  vector<int> ret;
  if (D[u] > D[v]) return lca(v, u);
  int diff = D[v] - D[u];
  for (int i = (0); i <= (16); i++)
    if (diff & (1 << i)) {
      merge(ret, P[v][i]);
      v = L[v][i];
    }
  if (u == v) {
    merge(ret, P[u][0]);
    return ret;
  }
  for (int i = (16); i >= (0); i--)
    if (L[u][i] != L[v][i]) {
      merge(ret, P[u][i]);
      u = L[u][i];
      merge(ret, P[v][i]);
      v = L[v][i];
    }
  merge(ret, P[u][0]);
  merge(ret, P[v][0]);
  merge(ret, P[L[v][0]][0]);
  return ret;
}
int main() {
  cin >> N >> M >> Q;
  for (int i = (1); i <= (N - 1); i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    E[a].push_back(b), E[b].push_back(a);
  }
  for (int i = (1); i <= (M); i++) {
    int c;
    scanf("%d", &c);
    P[c][0].push_back(i);
  }
  for (int i = (1); i <= (N); i++) {
    sort(P[i][0].begin(), P[i][0].end());
    while (((int)P[i][0].size()) > 10) P[i][0].pop_back();
  }
  dfs(1, 0, 1);
  for (int j = (1); j <= (16); j++)
    for (int i = (1); i <= (N); i++) {
      L[i][j] = L[L[i][j - 1]][j - 1];
      merge(P[i][j], P[L[i][j - 1]][j - 1]);
      merge(P[i][j], P[i][j - 1]);
    }
  while (Q--) {
    int u, v, a;
    scanf("%d %d %d", &u, &v, &a);
    vector<int> ans = lca(u, v);
    while (((int)ans.size()) > a) ans.pop_back();
    printf("%d ", ((int)ans.size()));
    for (int k : ans) printf("%d ", k);
    printf("\n");
  }
}