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 arr {
  int W[10];
};
arr MAX, C[4 * 200005];
vector<int> a[200005], V[200005];
int cnt[200005], T[200005], P[20][200005], pos[200005], num, B[200005],
    d[200005];
int n, m, q;
void DFS1(int u, int par) {
  cnt[u] = 1;
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (v == par) continue;
    T[v] = T[u] + 1;
    P[0][v] = u;
    DFS1(v, u);
    cnt[u] += cnt[v];
  }
}
void DFS2(int u, int par) {
  int Ma = 0;
  pos[u] = ++num;
  d[num] = u;
  if (B[u] == 0) B[u] = pos[u];
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (v == par) continue;
    if (cnt[Ma] < cnt[v] || Ma == 0) Ma = v;
  }
  if (Ma == 0) return;
  B[Ma] = B[u];
  DFS2(Ma, u);
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (v == par || v == Ma) continue;
    DFS2(v, u);
  }
}
int PAR(int u, int h) {
  int hi = T[u] - h;
  for (int i = 0; i < 20; i++)
    if (((hi >> i) & 1)) u = P[i][u];
  return u;
}
int LCA(int u, int v) {
  if (T[u] < T[v]) swap(u, v);
  u = PAR(u, T[v]);
  if (u == v) return u;
  for (int i = 19; i >= 0; i--)
    if (P[i][u] != P[i][v]) {
      u = P[i][u];
      v = P[i][v];
    }
  return P[0][u];
}
arr Merge(arr P, arr Q) {
  arr R;
  int pos1 = 0, pos2 = 0;
  for (int i = 0; i < 10; i++) {
    if (P.W[pos1] > Q.W[pos2]) {
      R.W[i] = Q.W[pos2];
      pos2++;
    } else {
      R.W[i] = P.W[pos1];
      pos1++;
    }
  }
  return R;
}
void Update(int node, int l, int r, int pos, int Val) {
  if (l == r && l == pos) {
    for (int i = 0; i < 10; i++)
      if (C[node].W[i] >= Val) swap(C[node].W[i], Val);
    return;
  }
  if (l > pos || r < pos) return;
  int g = (l + r) / 2;
  Update(node * 2, l, g, pos, Val);
  Update(node * 2 + 1, g + 1, r, pos, Val);
  C[node] = Merge(C[2 * node], C[2 * node + 1]);
}
arr Get_Val(int node, int l, int r, int d, int c) {
  if (d <= l && r <= c) return C[node];
  if (d > r || l > c) return MAX;
  int g = (l + r) / 2;
  arr u = Get_Val(node * 2, l, g, d, c);
  arr v = Get_Val(node * 2 + 1, g + 1, r, d, c);
  return Merge(u, v);
}
arr Get(int u, int par) {
  arr r = MAX;
  while (T[d[B[u]]] > T[par]) {
    r = Merge(r, Get_Val(1, 1, n, B[u], pos[u]));
    u = P[0][d[B[u]]];
  }
  r = Merge(r, Get_Val(1, 1, n, pos[par], pos[u]));
  return r;
}
int main() {
  cin >> n >> m >> q;
  int u, v, A;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    a[u].push_back(v);
    a[v].push_back(u);
  }
  for (int i = 0; i < 10; i++) MAX.W[i] = 1e9;
  DFS1(1, 0);
  for (int j = 1; j < 20; j++)
    for (int i = 1; i <= n; i++) P[j][i] = P[j - 1][P[j - 1][i]];
  DFS2(1, 0);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &u);
    V[u].push_back(i);
  }
  for (int i = 0; i < 4 * 200005; i++) C[i] = MAX;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < V[i].size(); j++) Update(1, 1, n, pos[i], V[i][j]);
  while (q--) {
    scanf("%d%d%d", &u, &v, &A);
    if (T[u] > T[v]) swap(u, v);
    int par = LCA(u, v);
    arr P = Get(u, par), Q = MAX;
    if (u != v) Q = Get(v, PAR(v, T[par] + 1));
    P = Merge(P, Q);
    for (int i = 9; i >= -1; i--) {
      if (i == -1) {
        A = -1;
        break;
      }
      if (P.W[i] != 1e9) {
        A = min(A - 1, i);
        break;
      }
    }
    printf("%d ", A + 1);
    for (int i = 0; i <= A; i++) printf("%d ", P.W[i]);
    printf("\n");
  }
}