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> g[101010];
vector<int> pep[101010][20];
int N;
int viz[101010];
int tata[101010][20];
int M, q;
int A = 20;
vector<int> combine(vector<int> &v1, vector<int> &v2) {
  vector<int> v(v1.size() + v2.size());
  merge(v1.begin(), v1.end(), v2.begin(), v2.end(), v.begin());
  v.erase(unique(v.begin(), v.end()), v.end());
  if (v.size() > A) v.resize(A);
  return v;
}
void dfs(int x, int t, int lev) {
  tata[x][0] = t;
  viz[x] = lev;
  for (auto n : g[x]) {
    if (!viz[n]) {
      dfs(n, x, lev + 1);
    }
  }
}
void afis(vector<int> &v) {
  cout << v.size() << endl;
  for (auto x : v) {
    cout << x << " ";
  }
  cout << endl;
}
vector<int> make(int x, int y) {
  if (viz[x] < viz[y]) {
    swap(x, y);
  }
  if (x == y) {
    return pep[x][0];
  }
  int log1, log2;
  for (log1 = 1; (1 << log1) < viz[x]; ++log1)
    ;
  for (log2 = 1; (1 << log2) < viz[y]; ++log2)
    ;
  vector<int> res = combine(pep[x][0], pep[y][0]);
  ;
  for (int k = log1; k >= 0; --k) {
    if (viz[x] - (1 << k) >= viz[y]) {
      res = combine(res, pep[x][k]);
      x = tata[x][k];
      res = combine(res, pep[x][0]);
    }
  }
  if (x == y) {
    res = combine(res, pep[x][0]);
    return res;
  }
  for (int k = log2; k >= 0; --k) {
    if (tata[x][k] && tata[x][k] != tata[y][k]) {
      res = combine(res, pep[x][k]);
      x = tata[x][k];
      res = combine(res, pep[x][0]);
      res = combine(res, pep[y][k]);
      y = tata[y][k];
      res = combine(res, pep[y][0]);
    }
  }
  return combine(res, pep[tata[x][0]][0]);
}
int main() {
  A = 20;
  scanf("%d%d%d", &N, &M, &q);
  for (int i = 1; i < N; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= M; ++i) {
    int x;
    scanf("%d", &x);
    pep[x][0].push_back(i);
  }
  for (int i = 1; i <= N; ++i) {
    sort(pep[i][0].begin(), pep[i][0].end());
    if (pep[i][0].size() > A) pep[i][0].resize(A);
  }
  dfs(1, 0, 1);
  for (int i = 1; i <= N; ++i) {
    --viz[i];
  }
  for (int k = 1; (1 << k) <= N; ++k) {
    for (int i = 1; i <= N; ++i) {
      tata[i][k] = tata[tata[i][k - 1]][k - 1];
      pep[i][k] = combine(pep[i][k - 1], pep[tata[i][k - 1]][k - 1]);
    }
  }
  for (int i = 1; i <= q; ++i) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    vector<int> v = make(x, y);
    printf("%d ", min((int)v.size(), z));
    for (int i = 0; i < min((int)v.size(), z); ++i) {
      printf("%d ", v[i]);
    }
    printf("\n");
  }
}