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 N = 100002, md = 1000000007, INF = 10000001;
using namespace std;
int n, l, x, y, m, q, A1, fe, P[N], D[21];
vector<vector<int> > g;
vector<int> tin, tout, pa, X[N], ans, ans1, mn[N][20];
int timer;
vector<vector<int> > up;
void dfs(int v, int p = 0) {
  tin[v] = ++timer;
  P[v] = p;
  up[v][0] = p;
  if (X[v].size())
    for (int i = 0; i < X[v].size(); i++) mn[v][0].push_back(X[v][i]);
  for (int i = 1; i <= l; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    merge(mn[v][i - 1].begin(), mn[v][i - 1].end(),
          mn[up[v][i - 1]][i - 1].begin(), mn[up[v][i - 1]][i - 1].end(),
          back_inserter(mn[v][i]));
    if (mn[v][i].size() > 10) mn[v][i].resize(10);
  }
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (to != p) dfs(to, v);
  }
  tout[v] = ++timer;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = l; i >= 0; --i)
    if (!upper(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
void get(int v, int p) {
  for (int i = l; i >= 0; --i)
    if (!upper(up[v][i], p)) {
      merge(mn[v][i].begin(), mn[v][i].end(), ans.begin(), ans.end(),
            back_inserter(ans1));
      ans = ans1;
      v = up[v][i];
      if (ans.size() > 10) ans.resize(10);
      ans1.clear();
    }
  if (p == P[v])
    if (X[v].size())
      for (int i = 0; i < X[v].size(); i++) ans.push_back(X[v][i]);
  sort(ans.begin(), ans.end());
  if (ans.size() > 10) ans.resize(10);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  tin.resize(n + 1), tout.resize(n + 1), up.resize(n + 1), g.resize(n + 1);
  l = 1;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &x);
    if (X[x].size() < 10) X[x].push_back(i);
  }
  while ((1 << l) <= n) ++l;
  for (int i = 0; i <= n; ++i) up[i].resize(l + 1);
  tin[0] = -123123;
  tout[0] = 21324324;
  dfs(1);
  while (q--) {
    scanf("%d%d%d", &x, &y, &A1);
    int lc = lca(x, y);
    ans.clear();
    get(x, lc);
    get(y, lc);
    int f = ans.size();
    if (X[lc].size()) {
      for (int i = 0; i < X[lc].size(); i++) ans.push_back(X[lc][i]);
    }
    sort(ans.begin(), ans.end());
    printf("%d ", min(A1, int(ans.size())));
    for (int i = 0; i < min(A1, int(ans.size())); i++) printf("%d ", ans[i]);
    puts("");
  }
}