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>
const int N = 1e5 + 10;
using namespace std;
struct ED {
  int v, next;
} e[N * 2];
int head[N], w[N], r[N], siz[N], z, EN;
int dep[N], fa[N], top[N], son[N];
struct tt {
  int z[20], c;
  tt() { c = 0; }
} T[N << 2];
int n, m, q, u, v, a;
void addedge(int u, int v) {
  e[EN] = ED{v, head[u]};
  head[u] = EN++;
}
void dfs(int u) {
  siz[u] = 1;
  son[u] = 0;
  for (int i = head[u]; ~i; i = e[i].next) {
    int v = e[i].v;
    if (!siz[v]) {
      fa[v] = u;
      dep[v] = dep[u] + 1;
      dfs(v);
      if (siz[v] > siz[son[u]]) son[u] = v;
      siz[u] += siz[v];
    }
  }
}
void build_tree(int u, int tp) {
  w[u] = ++z;
  r[z] = u;
  top[u] = tp;
  if (son[u]) build_tree(son[u], top[u]);
  for (int i = head[u]; ~i; i = e[i].next)
    if (!w[e[i].v]) build_tree(e[i].v, e[i].v);
}
tt pushup(tt &a, tt &b) {
  tt c;
  int i = 0, j = 0;
  while (c.c < 10 && i < a.c && j < b.c) {
    if (a.z[i] <= b.z[j])
      c.z[c.c++] = a.z[i++];
    else
      c.z[c.c++] = b.z[j++];
  }
  while (c.c < 10 && i < a.c) c.z[c.c++] = a.z[i++];
  while (c.c < 10 && j < b.c) c.z[c.c++] = b.z[j++];
  return c;
}
tt query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) {
    return T[rt];
  }
  int m = (l + r) / 2;
  tt a, b;
  if (L <= m) a = query(L, R, l, m, rt << 1);
  if (R > m) b = query(L, R, m + 1, r, rt << 1 | 1);
  return pushup(a, b);
}
tt Find(int u, int v) {
  int a = top[u], b = top[v];
  tt c, k;
  while (a != b) {
    if (dep[a] < dep[b]) swap(a, b), swap(u, v);
    k.c = 0;
    k = query(w[a], w[u], 1, n, 1);
    c = pushup(c, k);
    u = fa[a];
    a = top[u];
  }
  if (dep[u] > dep[v]) swap(u, v);
  k.c = 0;
  k = query(w[u], w[v], 1, n, 1);
  c = pushup(c, k);
  return c;
}
void init() {
  fa[1] = dep[1] = z = 0;
  memset(w, 0, sizeof(w));
  memset(siz, 0, sizeof(siz));
  dfs(1);
  build_tree(1, 1);
}
void build(int l, int r, int rt) {
  T[rt].c = 0;
  if (l == r) return;
  int m = (l + r) / 2;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void update(int L, int x, int l, int r, int rt) {
  if (L == l && l == r) {
    int f = -1;
    for (int i = 0; i < min(10, T[rt].c) && f == -1; i++)
      if (T[rt].z[i] > x) f = i;
    if (T[rt].c == 10) {
      if (f == -1) return;
      for (int i = 9; i >= f; i++) T[rt].z[i + 1] = T[rt].z[i];
      T[rt].z[f] = x;
    } else {
      if (f == -1)
        T[rt].z[T[rt].c++] = x;
      else {
        for (int i = 9; i >= f; i++) T[rt].z[i + 1] = T[rt].z[i];
        T[rt].z[f] = x;
      }
    }
    return;
  }
  int m = (l + r) / 2;
  if (L <= m)
    update(L, x, l, m, rt << 1);
  else
    update(L, x, m + 1, r, rt << 1 | 1);
  T[rt] = pushup(T[rt << 1], T[rt << 1 | 1]);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  memset(head, -1, sizeof(head));
  EN = 0;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  init();
  build(1, n, 1);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &a);
    update(w[a], i, 1, n, 1);
  }
  while (q--) {
    scanf("%d%d%d", &u, &v, &a);
    tt t = Find(u, v);
    printf("%d ", min(a, t.c));
    for (int i = 0; i < min(a, t.c); i++) printf("%d ", t.z[i]);
    printf("\n");
  }
  return 0;
}