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 M = 100005;
int h[M], tot;
int fa[M], son[M], top[M], sz[M], dep[M];
int n, m, q, mx, ans[M];
int id[M], R[M], num, S[15][M];
struct node {
  int to, nx;
} G[M << 1];
void Add(int a, int b) {
  G[++tot].to = b;
  G[tot].nx = h[a];
  h[a] = tot;
}
inline void Rd(int &x) {
  x = 0;
  char c;
  while (c = getchar(), !isdigit(c))
    ;
  do x = (x << 1) + (x << 3) + (c ^ 48);
  while (c = getchar(), isdigit(c));
}
void put(int x) {
  if (!x) return;
  put(x / 10);
  putchar(x % 10 ^ 48);
}
void pt(int x) {
  if (!x)
    putchar('0');
  else {
    if (x < 0) putchar('-'), x = -x;
    put(x);
  }
  putchar(' ');
}
void dfs(int x, int pre, int d) {
  sz[x] = 1;
  dep[x] = d;
  for (register int i = h[x]; i; i = G[i].nx) {
    int y = G[i].to;
    if (y == pre) continue;
    fa[y] = x;
    dfs(y, x, d + 1);
    sz[x] += sz[y];
    if (sz[y] > sz[son[x]]) son[x] = y;
  }
}
void rdfs(int x, int pre, int f) {
  top[x] = f;
  R[id[x] = ++num] = x;
  if (son[x]) rdfs(son[x], x, f);
  for (register int i = h[x]; i; i = G[i].nx) {
    int y = G[i].to;
    if (y == pre || y == son[x]) continue;
    rdfs(y, x, y);
  }
}
int T[22][M << 2];
void build(int l = 1, int r = n, int p = 1) {
  if (l == r) {
    for (int i = 1; i <= S[0][R[l]] && i <= 10; i++)
      T[++T[0][p]][p] = S[i][R[l]];
    return;
  }
  int x = 0, y = 0, mid = l + r >> 1;
  build(l, mid, p << 1);
  build(mid + 1, r, p << 1 | 1);
  while (T[0][p] < 10 && x < T[0][p << 1] && y < T[0][p << 1 | 1]) {
    if (T[x + 1][p << 1] < T[y + 1][p << 1 | 1])
      T[++T[0][p]][p] = T[++x][p << 1];
    else
      T[++T[0][p]][p] = T[++y][p << 1 | 1];
  }
  while (T[0][p] < 10 && x < T[0][p << 1]) T[++T[0][p]][p] = T[++x][p << 1];
  while (T[0][p] < 10 && y < T[0][p << 1 | 1])
    T[++T[0][p]][p] = T[++y][p << 1 | 1];
}
void query(int L, int R, int l = 1, int r = n, int p = 1) {
  if (r < L || R < l) return;
  if (L <= l && r <= R) {
    for (int i = 1; i <= min(T[0][p], mx); i++) ans[++ans[0]] = T[i][p];
    return;
  }
  int mid = l + r >> 1;
  query(L, R, l, mid, p << 1);
  query(L, R, mid + 1, r, p << 1 | 1);
}
void solve(int a, int b, int c) {
  mx = c;
  ans[0] = 0;
  while (top[a] != top[b]) {
    if (dep[top[a]] < dep[top[b]]) swap(a, b);
    query(id[top[a]], id[a]);
    a = fa[top[a]];
  }
  if (id[a] > id[b]) swap(a, b);
  query(id[a], id[b]);
  sort(ans + 1, ans + 1 + ans[0]);
  ans[0] = min(ans[0], mx);
  pt(ans[0]);
  for (int i = 1; i <= ans[0]; i++) pt(ans[i]);
  putchar('\n');
}
int main() {
  int a, b, c;
  Rd(n);
  Rd(m);
  Rd(q);
  for (int i = 1; i < n; i++) {
    Rd(a);
    Rd(b);
    Add(a, b);
    Add(b, a);
  }
  dfs(1, -1, 1);
  rdfs(1, -1, 1);
  for (int i = 1; i <= m; i++) {
    Rd(a);
    if (S[0][a] < 10) S[++S[0][a]][a] = i;
  }
  build();
  for (int i = 1; i <= q; i++) {
    Rd(a);
    Rd(b);
    Rd(c);
    solve(a, b, c);
  }
  return 0;
}