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> a[100100];
vector<int> d[100100];
vector<int> v[100100][22];
int T[100100][22];
int Niv[100100];
int niv[200100];
bool bf[100100];
int lim;
int rmq[200200][22], eul[200200], lg[200200], st[200200];
int nn = -1, n;
vector<int> interclasare(vector<int> v1, vector<int> v2) {
  vector<int> sol;
  int cn = 0;
  int i1 = 0, i2 = 0;
  while (cn < 10 && i1 < v1.size() && i2 < v2.size()) {
    if (v1[i1] <= v2[i2]) {
      sol.push_back(v1[i1]);
      cn++;
      i1++;
    } else {
      sol.push_back(v2[i2]);
      cn++;
      i2++;
    }
  }
  while (cn < 10 && i1 < v1.size()) {
    sol.push_back(v1[i1]);
    cn++;
    i1++;
  }
  while (cn < 10 && i2 < v2.size()) {
    sol.push_back(v2[i2]);
    cn++;
    i2++;
  }
  return sol;
}
void pre() {
  int i, j;
  lim = 0;
  int p = 1;
  while (p * 2 <= n) {
    p *= 2;
    lim++;
  }
  for (j = 1; j <= lim; j++) {
    for (i = 1; i <= n; i++) {
      int T1, T2;
      T1 = T[i][j - 1];
      if (T1 != -1) T1 = T[T1][1];
      if (T1 != -1) T2 = T[T1][j - 1];
      if (T1 != -1 && T2 != -1) {
        T[i][j] = T2;
        int cn = 0;
        int i1 = 0, i2 = 0;
        v[i][j] = interclasare(v[i][j - 1], v[T1][j - 1]);
      } else
        T[i][j] = -1;
    }
  }
}
void dfs(int x, int nv, int t) {
  int i;
  Niv[x] = nv;
  T[x][1] = t;
  bf[x] = true;
  eul[++nn] = x;
  st[x] = nn;
  niv[nn] = nv;
  for (i = 0; i < a[x].size(); i++)
    if (bf[a[x][i]] == false) {
      dfs(a[x][i], nv + 1, x);
      eul[++nn] = x;
      niv[nn] = nv;
    }
}
void pre_rmq() {
  int i, j;
  for (i = 2; i <= nn; i++) lg[i] = lg[i >> 1] + 1;
  for (i = 0; i <= nn; i++) rmq[i][0] = i;
  for (j = 1, i = 0; (1 << j) < nn; j++)
    for (i = 0; i <= nn - (1 << j); i++)
      if (niv[rmq[i][j - 1]] < niv[rmq[i + (1 << (j - 1))][j - 1]])
        rmq[i][j] = rmq[i][j - 1];
      else
        rmq[i][j] = rmq[i + (1 << (j - 1))][j - 1];
}
int get_lca(int x, int y) {
  int nr;
  x = st[x];
  y = st[y];
  if (x > y) {
    int ax = x;
    x = y;
    y = ax;
  }
  nr = lg[y - x + 1];
  if (niv[rmq[x][nr]] < niv[rmq[y - (1 << nr) + 1][nr]])
    return (eul[rmq[x][nr]]);
  else
    return (eul[rmq[y - (1 << nr) + 1][nr]]);
}
vector<int> query(int x, int y) {
  vector<int> sol1;
  vector<int> sol2;
  int i, j;
  int lca = get_lca(x, y);
  i = lim;
  while (true) {
    while (T[x][i] == -1 || Niv[T[x][i]] < Niv[lca]) i--;
    sol1 = interclasare(sol1, v[x][i]);
    x = T[x][i];
    if (x == lca) break;
    x = T[x][1];
  }
  i = lim;
  while (y != lca) {
    while (T[y][i] == -1 || Niv[T[y][i]] <= Niv[lca]) i--;
    sol2 = interclasare(sol2, v[y][i]);
    y = T[y][i];
    y = T[y][1];
  }
  return interclasare(sol1, sol2);
}
int main() {
  int m, q, i;
  scanf("%d%d%d", &n, &m, &q);
  for (i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  for (i = 1; i <= n; i++) T[i][0] = i;
  for (i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    v[x][0].push_back(i);
  }
  for (i = 1; i <= n; i++)
    if (v[i][0].size()) {
      sort(v[i][0].begin(), v[i][0].end());
      while (v[i][0].size() > 10) v[i][0].pop_back();
    }
  dfs(1, 0, -1);
  pre_rmq();
  pre();
  for (i = 1; i <= q; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    vector<int> s = query(x, y);
    printf("%d ", min(z, (int)s.size()));
    for (int j = 0; j < min(z, (int)s.size()); j++) printf("%d ", s[j]);
    printf("\n");
  }
  return 0;
}