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;
struct node {
  vector<int> humans;
  vector<int> to;
};
void merge_humans(vector<int> &x, vector<int> &y, vector<int> &ans) {
  vector<int> buffer = x;
  for (int i = 0; i < (int)y.size(); ++i) {
    buffer.push_back(y[i]);
  }
  sort(buffer.begin(), buffer.end());
  ans.clear();
  if (buffer.empty()) return;
  ans.push_back(buffer[0]);
  for (int i = 1; i < (int)buffer.size(); ++i) {
    if (buffer[i - 1] != buffer[i] && ans.size() < 10) {
      ans.push_back(buffer[i]);
    }
  }
}
struct LCA_node {
  vector<int> humans;
  int upper_node;
  LCA_node(){};
  LCA_node(std::vector<int> &a, std::vector<int> &b, int u) {
    merge_humans(a, b, humans);
    upper_node = u;
  }
  LCA_node(LCA_node &x, LCA_node &y) {
    upper_node = y.upper_node;
    merge_humans(x.humans, y.humans, humans);
  }
};
vector<node> g;
int n, m, q, tin[111111], tout[111111], timer;
const int depth = 23;
LCA_node lca[111111][depth];
void BuildLCA(int x, int y) {
  tin[x] = ++timer;
  lca[x][0] = LCA_node(g[x].humans, g[y].humans, y);
  for (int i = 1; i < depth; ++i) {
    lca[x][i] = LCA_node(lca[x][i - 1], lca[lca[x][i - 1].upper_node][i - 1]);
  }
  for (int i = 0; i < (int)g[x].to.size(); ++i) {
    if (g[x].to[i] != y) {
      BuildLCA(g[x].to[i], x);
    }
  }
  tout[x] = ++timer;
}
bool is_upper(int x, int y) { return tin[x] <= tin[y] && tout[x] >= tout[y]; }
int getLCA(int x, int y) {
  if (is_upper(x, y)) return x;
  if (is_upper(y, x)) return y;
  for (int i = depth - 1; i >= 0; --i) {
    if (!is_upper(lca[x][i].upper_node, y)) x = lca[x][i].upper_node;
  }
  return lca[x][0].upper_node;
}
void GetAns(int x, int y, int c) {
  vector<int> ans;
  int v_lca = getLCA(x, y);
  merge_humans(ans, g[x].humans, ans);
  merge_humans(ans, g[y].humans, ans);
  for (int i = depth - 1; i >= 0; --i) {
    if (!is_upper(lca[x][i].upper_node, v_lca)) {
      merge_humans(ans, lca[x][i].humans, ans);
      x = lca[x][i].upper_node;
    }
    if (!is_upper(lca[y][i].upper_node, v_lca)) {
      merge_humans(ans, lca[y][i].humans, ans);
      y = lca[y][i].upper_node;
    }
  }
  merge_humans(ans, g[v_lca].humans, ans);
  sort(ans.begin(), ans.end());
  printf("%d ", min(c, (int)ans.size()));
  for (int i = 0; (int)i < min(c, (int)ans.size()); ++i) {
    printf("%d ", ans[i]);
  }
  printf("\n");
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  g.resize(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].to.push_back(y);
    g[y].to.push_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    int x;
    scanf("%d", &x);
    g[x].humans.push_back(i);
  }
  for (int i = 0; i <= n; ++i) {
    sort(g[i].humans.begin(), g[i].humans.end());
    if (g[i].humans.size() > 10) {
      g[i].humans.resize(10);
    }
  }
  BuildLCA(1, 0);
  tout[0] = ++timer;
  for (int i = 0; i < q; ++i) {
    int v, u, a;
    scanf("%d %d %d", &v, &u, &a);
    GetAns(v, u, a);
  }
  return 0;
}