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[100010];
int t[100010], p[18][100010], L[100010], d[100010];
vector<int> a[18][100010], c[100010];
void dfs(int u) {
  for (int i = 0; i < (int)g[u].size(); i++) {
    int v = g[u][i];
    if (t[v] == -1) {
      t[v] = u;
      d[v] = d[u] + 1;
      dfs(v);
    }
  }
}
void Merge(const vector<int>& A, const vector<int>& B, vector<int>& C,
           int L = 10) {
  int i = 0, j = 0, k = 0, n = (int)A.size(), m = (int)B.size();
  while (i < n && j < m && k < L) {
    if (A[i] < B[j]) {
      C.push_back(A[i]);
      i++, k++;
    } else {
      C.push_back(B[j]);
      j++, k++;
    }
  }
  while (i < n && k < L) {
    C.push_back(A[i]);
    i++, k++;
  }
  while (j < m && k < L) {
    C.push_back(B[j]);
    j++, k++;
  }
}
int scan() {
  int ret = 0;
  int ip = getchar();
  for (; ip < '0' || ip > '9'; ip = getchar())
    ;
  for (; ip >= '0' && ip <= '9'; ip = getchar()) ret = ret * 10 + ip - '0';
  return ret;
}
int main() {
  int n, m, q, i, j, u, v, l, x, y;
  vector<int> ans, temp, temp1;
  L[1] = 0;
  for (i = 2; i < 100010; i++) L[i] = L[i >> 1] + 1;
  n = scan();
  m = scan();
  q = scan();
  for (i = 0; i < n - 1; i++) {
    u = scan();
    v = scan();
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (i = 0; i < m; i++) {
    u = scan();
    --u;
    c[u].push_back(i + 1);
  }
  for (i = 0; i < n; i++) t[i] = -1;
  t[0] = 0;
  d[0] = 0;
  dfs(0);
  t[0] = -1;
  for (i = 0; i < n; i++)
    for (j = 0; (1 << j) < n; j++) p[j][i] = -1;
  for (i = 0; i < n; i++) {
    p[0][i] = t[i];
    if (t[i] != -1 && (int)c[t[i]].size() > 0)
      for (j = 0; j < (int)c[t[i]].size() && j < 10; j++)
        a[0][i].push_back(c[t[i]][j]);
  }
  for (j = 1; (1 << j) < n; j++) {
    for (i = 0; i < n; i++) {
      if (p[j - 1][i] != -1) {
        p[j][i] = p[j - 1][p[j - 1][i]];
        Merge(a[j - 1][i], a[j - 1][p[j - 1][i]], a[j][i]);
      }
    }
  }
  while (q--) {
    u = scan();
    v = scan();
    m = scan();
    --u, --v;
    if (u == v) {
      if (!c[u].empty()) {
        m = min(m, (int)c[u].size());
        printf("%d", m);
        for (i = 0; i < m; i++) printf(" %d", c[u][i]);
        printf("\n");
      } else
        printf("0\n");
      continue;
    }
    x = u;
    y = v;
    temp.clear();
    temp1.clear();
    ans.clear();
    if (d[u] < d[v]) swap(u, v);
    l = L[d[u]];
    for (i = l; i >= 0; i--)
      if (d[u] - (1 << i) >= d[v]) {
        temp = ans;
        ans.clear();
        Merge(temp, a[i][u], ans, m);
        u = p[i][u];
      }
    if (u != v) {
      for (i = l; i >= 0; i--)
        if (p[i][u] != -1 && p[i][u] != p[i][v]) {
          temp.clear();
          Merge(a[i][u], a[i][v], temp, m);
          temp1 = ans;
          ans.clear();
          Merge(temp, temp1, ans, m);
          u = p[i][u];
          v = p[i][v];
        }
      temp = ans;
      ans.clear();
      Merge(temp, a[0][u], ans, m);
    }
    set<int> s;
    s.clear();
    for (i = 0; i < m && i < (int)c[x].size(); i++) s.insert(c[x][i]);
    for (i = 0; i < m && i < (int)c[y].size(); i++) s.insert(c[y][i]);
    for (i = 0; i < (int)ans.size(); i++) s.insert(ans[i]);
    ans.clear();
    i = 0;
    for (set<int>::iterator it = s.begin(); i < m && it != s.end(); it++, i++)
      if (*it != 0) ans.push_back(*it);
    m = min(m, (int)ans.size());
    printf("%d", m);
    for (i = 0; i < m; i++) printf(" %d", ans[i]);
    printf("\n");
  }
  return 0;
}