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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2")
using namespace std;
const int mxn = 100 * 1000 + 3, lgn = 18;
int dt[mxn], up[mxn][lgn];
vector<int> g[mxn], par[mxn][lgn];
int n, m, q;
vector<int> merge(vector<int> f, vector<int> gg) {
  if (f.size() < gg.size()) swap(f, gg);
  for (int x : gg) f.push_back(x);
  sort(f.begin(), f.end());
  while (f.size() > 10) f.pop_back();
  return f;
}
void dfs(int v, int p = 1) {
  up[v][0] = p;
  for (int i = 1; i < lgn; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    par[v][i] = merge(par[v][i - 1], par[up[v][i - 1]][i - 1]);
  }
  for (int u : g[v])
    if (u != p) {
      dt[u] = dt[v] + 1;
      dfs(u, v);
    }
}
int kth(int a, int diff) {
  for (int i = 0; i < lgn; i++)
    if (diff >> i & 1) a = up[a][i];
  return a;
}
int lca(int a, int b) {
  if (dt[a] > dt[b]) swap(a, b);
  b = kth(b, dt[b] - dt[a]);
  if (a == b) return a;
  for (int i = lgn - 1; i >= 0; i--)
    if (up[a][i] != up[b][i]) a = up[a][i], b = up[b][i];
  return up[a][0];
}
vector<int> fix(int v, int d) {
  vector<int> ret;
  for (int i = 0; i < lgn; i++) {
    if (d >> i & 1) {
      ret = merge(ret, par[v][i]);
      v = up[v][i];
    }
  }
  return ret;
}
signed main() {
  std::ios::sync_with_stdio(0);
  std::cout.tie(0);
  std::cin.tie(0);
  cin >> n >> m >> q;
  for (int u, v, i = 1; i < n; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int c, i = 1; i <= m; i++) {
    cin >> c;
    par[c][0].push_back(i);
    sort(par[c][0].begin(), par[c][0].end());
    while (par[c][0].size() > 10) par[c][0].pop_back();
  }
  dfs(1);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    int l = lca(u, v);
    vector<int> ans = merge(fix(u, dt[u] - dt[l]), fix(v, dt[v] - dt[l]));
    ans = merge(ans, par[l][0]);
    while (ans.size() > a) ans.pop_back();
    cout << ans.size() << ' ';
    for (int x : ans) cout << x << ' ';
    cout << '\n';
  }
  return 0;
}