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 = 1e5 + 5;
vector<int> c[N];
vector<int> s[20][N];
int dp[20][N];
int h[N];
vector<int> v[N];
int sh(int x, int y) {
  for (int i = 17; i + 1; --i)
    if (y & (1 << i)) x = dp[i][x];
  return x;
}
int ac(int x, int y) {
  if (h[x] > h[y])
    x = sh(x, h[x] - h[y]);
  else
    y = sh(y, h[y] - h[x]);
  for (int i = 17; i + 1 && x != y; --i)
    if (dp[i][x] != dp[i][y]) x = dp[i][x], y = dp[i][y];
  if (x == y)
    return x;
  else
    return dp[0][x];
}
void dfs(int prec, int node) {
  h[node] = h[prec] + 1;
  dp[0][node] = prec;
  s[0][node] = c[node];
  for (auto it : v[node])
    if (it != prec) dfs(node, it);
}
vector<int> get(vector<int> a, vector<int> b) {
  if (a.empty()) return b;
  if (b.empty()) return a;
  int *p = new int[a.size() + b.size()];
  merge(a.begin(), a.end(), b.begin(), b.end(), p);
  vector<int> as;
  for (int *r = p, cn = 1; cn <= a.size() + b.size() && cn <= 10; ++r, ++cn)
    as.push_back(*r);
  while (as.size() && !(*as.begin())) as.erase(as.begin());
  return as;
}
int main(void) {
  int n, m, q;
  scanf("%d%d%d\n", &n, &m, &q);
  int a, b;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &a);
    c[a].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    sort(c[i].begin(), c[i].end());
    while (c[i].size() > 10) c[i].erase(c[i].end() - 1);
  }
  dfs(0, 1);
  for (int i = 1; i < 18; ++i)
    for (int j = 1; j <= n; ++j) {
      dp[i][j] = dp[i - 1][dp[i - 1][j]];
      s[i][j] = get(s[i - 1][j], s[i - 1][dp[i - 1][j]]);
    }
  while (q--) {
    int a, b, mx;
    scanf("%d%d%d", &a, &b, &mx);
    int anc = ac(a, b);
    int disa = h[a] - h[anc];
    int disb = h[b] - h[anc];
    vector<int> ans;
    for (int i = 17; i + 1; --i)
      if ((1 << i) & disa) {
        ans = get(ans, s[i][a]);
        a = dp[i][a];
        while (ans.size() > mx) ans.erase(ans.end() - 1);
      }
    for (int i = 17; i + 1; --i)
      if ((1 << i) & disb) {
        ans = get(ans, s[i][b]);
        b = dp[i][b];
        while (ans.size() > mx) ans.erase(ans.end() - 1);
      }
    ans = get(ans, c[anc]);
    while (ans.size() > mx) ans.erase(ans.end() - 1);
    printf("%d ", ans.size());
    for (auto it : ans) printf("%d ", it);
    printf("\n");
  }
  return 0;
}