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;
int par[26][100005], depth[100005];
vector<int> st[400005];
vector<int> G[100005];
vector<int> SS[26][100005];
int cnt, chaincnt;
void get10(vector<int> &res, const vector<int> &fir, const vector<int> sec) {
  int c1 = 0, c2 = 0;
  res = vector<int>(min((int)fir.size() + (int)sec.size(), 10));
  for (int i = 0; i < (int)res.size(); i++) {
    if (c1 != fir.size() && c2 != sec.size()) {
      if (fir[c1] < sec[c2]) {
        res[i] = fir[c1++];
      } else {
        res[i] = sec[c2++];
      }
    } else if (c1 == fir.size()) {
      res[i] = sec[c2++];
    } else {
      res[i] = fir[c1++];
    }
  }
  return;
}
set<int> ret1, ret2;
int LCA(int u, int v) {
  ret1.clear(), ret2.clear();
  if (depth[u] < depth[v]) swap(u, v);
  int dif = depth[u] - depth[v];
  for (int i = 0; i <= 24; i++) {
    if (dif & (1 << i)) {
      u = par[i][u];
    }
  }
  if (u == v) return u;
  for (int i = 24; i >= 0; i--) {
    if (par[i][u] != par[i][v]) {
      u = par[i][u], v = par[i][v];
    }
  }
  return par[0][u];
}
vector<int> solve(int u, int v) {
  int lca = LCA(u, v);
  int d1 = depth[u] - depth[lca] + 1;
  int d2 = depth[v] - depth[lca] + 1;
  int lvl = 26;
  set<int> ret1;
  while (d1 > 0) {
    if ((1 << lvl) <= d1) {
      d1 -= (1 << lvl);
      for (int i = 0; i < (int)SS[lvl][u].size() && i < 10; i++) {
        ret1.insert(SS[lvl][u][i]);
      }
      while (ret1.size() > 10) {
        ret1.erase(--ret1.end());
      }
      u = par[lvl][u];
    }
    lvl--;
    if (lvl < 0) lvl = 0;
  }
  lvl = 26;
  set<int> ret2;
  while (d2 > 0) {
    if ((1 << lvl) <= d2) {
      d2 -= (1 << lvl);
      for (int i = 0; i < (int)SS[lvl][v].size() && i < 10; i++) {
        ret2.insert(SS[lvl][v][i]);
      }
      while (ret2.size() > 10) {
        ret2.erase(--ret2.end());
      }
      v = par[lvl][v];
    }
    lvl--;
    if (lvl < 0) lvl = 0;
  }
  ret1.insert(ret2.begin(), ret2.end());
  while (ret1.size() > 10) {
    ret1.erase(--ret1.end());
  }
  vector<int> ret(ret1.begin(), ret1.end());
  return ret;
}
void dfs(int u, int p, int dep) {
  depth[u] = dep;
  par[0][u] = p;
  for (int i = 0; i < (int)G[u].size(); i++) {
    int v = G[u][i];
    if (v == p) continue;
    dfs(v, u, dep + 1);
  }
}
int main() {
  int from, to;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &from, &to);
    G[from].push_back(to);
    G[to].push_back(from);
  }
  int cur;
  for (int i = 1; i <= m; i++) {
    scanf("%d", &cur);
    SS[0][cur].push_back(i);
  }
  memset(par, -1, sizeof par);
  chaincnt = cnt = 1;
  dfs(1, -1, 0);
  for (int i = 1; i <= 24; i++)
    for (int j = 1; j <= n; j++)
      if (par[i - 1][j] != -1) {
        par[i][j] = par[i - 1][par[i - 1][j]];
        get10(SS[i][j], SS[i - 1][par[i - 1][j]], SS[i - 1][j]);
      }
  int how;
  vector<int> ans;
  while (q--) {
    scanf("%d %d %d", &from, &to, &how);
    ans = solve(from, to);
    printf("%d", min((int)ans.size(), how));
    for (int i = 0; i < ans.size() && how; i++, how--) {
      printf(" %d", ans[i]);
    }
    printf("\n");
  }
  return 0;
}