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 = 100001;
int In[N];
int Out[N];
int t = 0;
int Euler[N << 3];
int Levels[N << 3];
class SparseTable {
 public:
  int* l2;
  int* M[30];
  int getMinIndex(int i, int j) {
    int k = l2[j - i + 1];
    if (A[M[k][i]] <= A[M[k][j - (1 << k) + 1]]) return M[k][i];
    return M[k][j - (1 << k) + 1];
  }
  int getMinVal(int i, int j) { return A[getMinIndex(i, j)]; }
  SparseTable(int t) {
    A.assign(t, 0);
    l2 = new int[t * 2];
    for (int i = 0; i < t; i++) {
      A[i] = Levels[i];
    }
    l2[1] = 0;
    l2[2] = 1;
    for (int i = 0; i < 30; i++) {
      M[i] = new int[t];
    }
    for (int i = 3; i < 2 * t; i++) {
      if ((i & (i - 1)) == 0)
        l2[i] = l2[i - 1] + 1;
      else
        l2[i] = l2[i - 1];
    }
    n = A.size();
    int i, j;
    for (i = 0; i < n; i++) M[0][i] = i;
    for (j = 1; 1 << j <= n; j++) {
      for (i = 0; i + (1 << j) - 1 < n; i++) {
        if (A[M[j - 1][i]] < A[M[j - 1][i + (1 << (j - 1))]])
          M[j][i] = M[j - 1][i];
        else
          M[j][i] = M[j - 1][i + (1 << (j - 1))];
      }
    }
  }

 private:
  int n;
  vector<int> A;
};
struct st {
  int arr[10];
  int sz;
  st() {
    sz = 0;
    for (int i = 0; i < 10; i++) {
      arr[i] = INT_MAX;
    }
  }
  st(int a) {
    for (int i = 0; i < 10; i++) {
      arr[i] = INT_MAX;
    }
    arr[0] = a;
    sz = 1;
  }
  st operator+(const st& b) {
    st z;
    int filled = 0, ap = 0, cp = 0;
    int tofill = min(sz + b.sz, 10);
    while (filled != tofill) {
      if (arr[ap] < b.arr[cp]) {
        z.arr[filled++] = arr[ap++];
      } else {
        z.arr[filled++] = b.arr[cp++];
      }
    }
    z.sz = filled;
    return z;
  }
};
st nodeData[N];
int parent[N];
int num[N];
int L[N];
int vis[N];
st part[N][20];
vector<int> centroidTree[N];
vector<int> adj[N];
void addEdge(int u, int v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
int sz[N];
void dfs(int s, int p = -1) {
  sz[s] = 1;
  for (auto c : adj[s]) {
    if (c == p || vis[c]) continue;
    dfs(c, s);
    sz[s] += sz[c];
  }
}
void dfsData(int s, int f, int p = -1, st cont = st()) {
  part[s][L[s] - L[f]] = cont;
  for (auto c : adj[s]) {
    if (c != p && (!vis[c])) {
      dfsData(c, f, s, cont + nodeData[c]);
    }
  }
}
int findCentroid(int s) {
  dfs(s, -1);
  int total = sz[s];
  int p = -1;
  for (int big = -1;; big = -1) {
    for (auto c : adj[s]) {
      if (c != p && (!vis[c])) {
        if (big == -1 || sz[big] < sz[c]) big = c;
      }
    }
    if (big == -1 || sz[big] * 2 <= total)
      break;
    else {
      p = s;
      s = big;
    }
  }
  return s;
}
int createTree(int s, int d = 0) {
  int ce = findCentroid(s);
  Euler[t] = ce;
  Levels[t] = d;
  In[ce] = t++;
  vis[ce] = true;
  L[ce] = d;
  for (auto ch : adj[ce]) {
    if (!vis[ch]) {
      int x = createTree(ch, d + 1);
      Euler[t] = ce;
      Levels[t] = d;
      Out[ce] = t++;
      parent[x] = ce;
      centroidTree[ce].push_back(x);
    }
  }
  dfsData(ce, ce, -1, st());
  vis[ce] = false;
  Euler[t] = ce;
  Levels[t] = d;
  Out[ce] = t++;
  return ce;
}
SparseTable* S;
int lca(int u, int v) {
  int mn = min(In[u], In[v]);
  int mx = max(Out[u], Out[v]);
  return Euler[S->getMinIndex(min(In[u], In[v]), max(Out[u], Out[v]))];
}
st query(int u, int v) {
  int _lca = lca(u, v);
  return part[u][L[u] - L[_lca]] + part[v][L[v] - L[_lca]] + nodeData[_lca];
}
void print(st a, int k) {
  int c = 0;
  for (int i = 0; i < 10; i++) {
    if (a.arr[i] != INT_MAX) c++;
  }
  c = min(c, k);
  printf("%d ", c);
  for (int i = 0; i < c; i++) {
    printf("%d ", a.arr[i]);
  }
  printf("\n");
}
int main() {
  srand(time(NULL));
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    nodeData[x] = nodeData[x] + st(i);
  }
  createTree(1);
  S = new SparseTable(t);
  while (q--) {
    int u, v, k;
    scanf("%d%d%d", &u, &v, &k);
    print(query(u, v), k);
  }
  return 0;
}