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>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 5;
const int D = 20;
int k = 1, head[maxn], fa[maxn][D + 2], n, m, q, d[maxn];
struct edge {
  int v, nxt;
} e[maxn * 2];
struct node {
  int d[15], size;
  node() {
    memset(d, 0, sizeof(d));
    size = 0;
  }
  void add(int num) {
    int pos = lower_bound(d, d + size, num) - d;
    for (int i = size; i > pos; i--) d[i] = d[i - 1];
    d[pos] = num;
    size = min(size + 1, 10);
  }
} ans[maxn][D + 2];
node operator+(const node &a, const node &b) {
  node c;
  int i = 0, j = 0, k = 0;
  while (i < a.size && j < b.size && k <= 10)
    if (a.d[i] < b.d[j])
      c.d[k++] = a.d[i++];
    else
      c.d[k++] = b.d[j++];
  while (i < a.size && k <= 10) c.d[k++] = a.d[i++];
  while (j < b.size && k <= 10) c.d[k++] = b.d[j++];
  c.size = k;
  return c;
}
void adds(int u, int v) {
  e[k].v = v;
  e[k].nxt = head[u];
  head[u] = k++;
}
void dfs(int u, int f) {
  d[u] = d[f] + 1;
  fa[u][0] = f;
  for (int i = 1; i <= D; ++i) {
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
    ans[u][i] = ans[fa[u][i - 1]][i - 1] + ans[u][i - 1];
  }
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].v != f) dfs(e[i].v, u);
}
node lca(int u, int v) {
  node s;
  if (u == v) return s + ans[u][0];
  if (d[u] < d[v]) swap(u, v);
  for (int i = D; i >= 0; --i)
    if (d[v] <= d[fa[u][i]]) {
      s = s + ans[u][i];
      u = fa[u][i];
    }
  if (u == v) return s + ans[u][0];
  for (int i = D; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) {
      s = s + ans[u][i];
      s = s + ans[v][i];
      u = fa[u][i];
      v = fa[v][i];
    }
  s = s + ans[u][0];
  s = s + ans[v][0];
  s = s + ans[fa[u][0]][0];
  return s;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adds(u, v);
    adds(v, u);
  }
  for (int i = 1; i <= m; ++i) {
    int a;
    scanf("%d", &a);
    ans[a][0].add(i);
  }
  dfs(1, 0);
  for (int i = 1; i <= q; ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    node s = lca(a, b);
    int t = min(s.size, c);
    printf("%d", t);
    for (int i = 0; i < t; ++i) printf(" %d", s.d[i]);
    printf("\n");
  }
  return 0;
}
