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;
vector<int> e[100005];
int c[100005];
vector<int> mn[20][100005];
int fa[20][100005], dep[100005];
int n, m, q;
void dfs(int u, int father, int d) {
  dep[u] = d;
  fa[0][u] = father;
  for (auto v : e[u]) {
    if (v == father) continue;
    dfs(v, u, d + 1);
  }
}
void initLCA() {
  int i, j;
  for (i = 1; i < 20; i++) {
    for (j = 1; j <= n; j++) {
      mn[i][j].resize(20);
      int k = fa[i - 1][j];
      fa[i][j] = fa[i - 1][k];
      merge(mn[i - 1][k].begin(), mn[i - 1][k].end(), mn[i - 1][j].begin(),
            mn[i - 1][j].end(), mn[i][j].begin());
      while (mn[i][j].size() > 10 || *(mn[i][j].end() - 1) == 0)
        mn[i][j].pop_back();
    }
  }
}
void merging(vector<int> &x, vector<int> &y) {
  vector<int> tmp(20);
  merge(x.begin(), x.end(), y.begin(), y.end(), tmp.begin());
  while (tmp.size() > 10 || *(tmp.end() - 1) == 0) tmp.pop_back();
  swap(x, tmp);
}
void queryLCA(int x, int y, vector<int> &a) {
  int i, j, dif = abs(dep[x] - dep[y]);
  if (dep[x] < dep[y]) swap(x, y);
  for (i = 20 - 1; i >= 0; i--) {
    if (dif & (1 << i)) {
      merging(a, mn[i][x]);
      x = fa[i][x];
    }
  }
  for (i = 20 - 1; i >= 0; i--) {
    if (fa[i][x] != fa[i][y]) {
      merging(a, mn[i][x]);
      x = fa[i][x];
      merging(a, mn[i][y]);
      y = fa[i][y];
    }
  }
  merging(a, mn[0][x]);
  if (x != y) {
    merging(a, mn[0][y]);
    merging(a, mn[0][fa[0][x]]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  while (cin >> n >> m >> q) {
    int x, y, z;
    for (int i = (1); i <= (n - 1); i++) {
      cin >> x >> y;
      e[x].push_back(y);
      e[y].push_back(x);
    }
    for (int i = (1); i <= (m); i++) {
      cin >> x;
      if (mn[0][x].size() < 10) mn[0][x].push_back(i);
    }
    for (int i = (0); i <= (20 - 1); i++) {
      for (int j = (1); j <= (n); j++) {
        sort(mn[i][j].begin(), mn[i][j].end());
      }
    }
    fa[0][0] = 0;
    dfs(1, 0, 0);
    initLCA();
    vector<int> ans;
    for (int i = (1); i <= (q); i++) {
      cin >> x >> y >> z;
      ans.clear();
      queryLCA(x, y, ans);
      z = ans.size() < z ? ans.size() : z;
      printf("%d ", z);
      for (int i = (0); i <= (z - 1); i++) printf("%d ", ans[i]);
      printf("\n");
    }
  }
  return 0;
}