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;
int N, M, Q;
vector<int> adj[100010];
int depth[100010];
int liftVert[100010][20], liftDP[100010][20][10];
void print(int A[10]) {
  for (int i = 0; i < 10; i++) cout << A[i] << " ";
  cout << "\n";
}
void DFS(int X, int P) {
  liftVert[X][0] = P;
  for (int NXT : adj[X]) {
    if (NXT != P) {
      depth[NXT] = depth[X] + 1;
      DFS(NXT, X);
    }
  }
}
void insert(int A[10], int X) {
  int pos = -1;
  for (int i = 0; i < 10; i++) {
    if (A[i] > X) {
      pos = i;
      break;
    }
  }
  if (pos != -1) {
    for (int i = 9; i > pos; i--) {
      A[i] = A[i - 1];
    }
    A[pos] = X;
  }
}
void merge(int res[10], int A[10], int B[10]) {
  int L = 0;
  int R = 0;
  int ind = 0;
  while (ind < 10) {
    if (R == 10 || A[L] <= B[R]) {
      res[ind++] = A[L++];
    } else if (L == 10 || B[R] < A[L]) {
      res[ind++] = B[R++];
    }
  }
}
void merge(int res[10], int A[10]) {
  int tmp[10];
  memset(tmp, 0, sizeof tmp);
  merge(tmp, res, A);
  for (int i = 0; i < 10; i++) res[i] = tmp[i];
}
void LCA(int res[10], int A, int B) {
  if (depth[A] < depth[B]) swap(A, B);
  int D = depth[A] - depth[B];
  for (int i = 0; i < 20; i++) {
    if (D & (1 << i)) {
      merge(res, liftDP[A][i]);
      A = liftVert[A][i];
    }
  }
  if (A == B) {
    merge(res, liftDP[A][0]);
    return;
  }
  for (int i = 19; i >= 0; i--) {
    if (liftVert[A][i] != liftVert[B][i]) {
      merge(res, liftDP[A][i]);
      merge(res, liftDP[B][i]);
      A = liftVert[A][i], B = liftVert[B][i];
    }
  }
  merge(res, liftDP[A][0]);
  merge(res, liftDP[B][0]);
  merge(res, liftDP[liftVert[A][0]][0]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M >> Q;
  for (int i = 1; i < N; i++) {
    int A, B;
    cin >> A >> B;
    adj[A].push_back(B), adj[B].push_back(A);
  }
  memset(liftDP, 0x3f, sizeof liftDP);
  for (int i = 1; i <= M; i++) {
    int C;
    cin >> C;
    insert(liftDP[C][0], i);
  }
  DFS(1, 0);
  for (int j = 1; j < 20; j++) {
    for (int i = 1; i <= N; i++) {
      liftVert[i][j] = liftVert[liftVert[i][j - 1]][j - 1];
      merge(liftDP[i][j], liftDP[i][j - 1], liftDP[liftVert[i][j - 1]][j - 1]);
    }
  }
  for (int i = 0; i < Q; i++) {
    int A, B, S;
    cin >> A >> B >> S;
    int cur[10];
    memset(cur, 0x3f, sizeof cur);
    LCA(cur, A, B);
    int sz = 0;
    vector<int> res;
    while (sz < S && cur[sz] != 0x3f3f3f3f) {
      res.push_back(cur[sz++]);
    }
    cout << sz << " ";
    for (int i : res) cout << i << " ";
    cout << "\n";
  }
}