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;
typedef int city[11];
int nex[100005 << 1], des[100005 << 1], head[100005];
city ans, tr[100005 << 2], c[100005];
int x, y, ll, rr, n, m, Q, k, top, tot, tid[100005], fa[100005], up[100005],
    sz[100005], son[100005], dep[100005];
void link(int x, int y) {
  nex[++top] = head[x];
  head[x] = top;
  des[top] = y;
}
void ins(city C, int num) {
  int br = 0;
  for (int i = 1; i <= C[0]; i++)
    if (num > C[i]) br = i;
  if (br == 10) return;
  for (int i = min(C[0], 9); i > br; i--) C[i + 1] = C[i];
  C[br + 1] = num;
  C[0] = min(10, C[0] + 1);
}
void add(city c1, city c2, city &c3) {
  city tmp;
  memset(tmp, 0, sizeof(tmp));
  int i = c1[0] ? 1 : 0, j = c2[0] ? 1 : 0;
  while (tmp[0] < 10) {
    if (i != 0 && j != 0) {
      tmp[++tmp[0]] = min(c1[i], c2[j]);
      if (tmp[tmp[0]] == c1[i]) {
        i++;
        if (i > c1[0]) i = 0;
      } else {
        j++;
        if (j > c2[0]) j = 0;
      }
    } else if (i == 0 && j == 0)
      break;
    else if (i == 0) {
      tmp[++tmp[0]] = c2[j++];
      if (j > c2[0]) j = 0;
    } else {
      tmp[++tmp[0]] = c1[i++];
      if (i > c1[0]) i = 0;
    }
  }
  for (int i = 0; i < 11; i++) c3[i] = tmp[i];
}
void dfs1(int now, int las) {
  fa[now] = las;
  sz[now] = 1;
  dep[now] = dep[las] + 1;
  int maxs = 0, maxx = 0;
  for (int i = head[now]; i; i = nex[i])
    if (des[i] != las) {
      dfs1(des[i], now);
      sz[now] += sz[des[i]];
      if (sz[des[i]] > maxx) maxx = sz[des[i]], maxs = des[i];
    }
  son[now] = maxs;
}
void dfs2(int now, int ant) {
  tid[now] = ++tot;
  up[now] = ant;
  if (son[now]) dfs2(son[now], ant);
  for (int i = head[now]; i; i = nex[i])
    if (!tid[des[i]]) dfs2(des[i], des[i]);
}
void build(int tt, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 11; i++) tr[tt][i] = c[l][i];
    return;
  }
  int mid = (l + r) >> 1;
  build(tt << 1, l, mid);
  build((tt << 1) + 1, mid + 1, r);
  add(tr[tt << 1], tr[(tt << 1) + 1], tr[tt]);
}
void ask(int tt, int l, int r) {
  if (ll <= l && rr >= r) {
    add(tr[tt], ans, ans);
    return;
  };
  int mid = (l + r) >> 1;
  if (ll <= mid) ask(tt << 1, l, mid);
  if (rr > mid) ask((tt << 1) + 1, mid + 1, r);
}
int main() {
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i < n; i++) scanf("%d%d", &x, &y), link(x, y), link(y, x);
  dfs1(1, 0);
  dfs2(1, 1);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &x);
    ins(c[tid[x]], i);
  }
  build(1, 1, n);
  while (Q--) {
    scanf("%d%d%d", &x, &y, &k);
    memset(ans, 0, sizeof(ans));
    while (up[x] != up[y]) {
      if (dep[up[x]] < dep[up[y]]) swap(x, y);
      ll = tid[up[x]], rr = tid[x];
      ask(1, 1, n);
      x = fa[up[x]];
    }
    if (dep[x] < dep[y]) swap(x, y);
    ll = tid[y], rr = tid[x];
    ask(1, 1, n);
    int ed = min(k, ans[0]);
    printf("%d ", ed);
    for (int i = 1; i <= ed; i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}