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;
void bug(string m = "here") { cout << m << endl; }
template <typename __ll>
inline void READ(__ll &m) {
  __ll x = 0, f = 1;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  m = x * f;
}
template <typename __ll>
inline void read(__ll &m) {
  READ(m);
}
template <typename __ll>
inline void read(__ll &m, __ll &a) {
  READ(m);
  READ(a);
}
template <typename __ll>
inline void read(__ll &m, __ll &a, __ll &b) {
  READ(m);
  READ(a);
  READ(b);
}
template <typename __ll>
inline void read(__ll &m, __ll &a, __ll &b, __ll &c) {
  READ(m);
  READ(a);
  READ(b);
  READ(c);
}
template <typename __ll>
inline void read(__ll &m, __ll &a, __ll &b, __ll &c, __ll &d) {
  READ(m);
  READ(a);
  READ(b);
  READ(c);
  read(d);
}
template <class T>
inline void out(T a) {
  if (a < 0) {
    putchar('-');
    a = -a;
  }
  if (a > 9) out(a / 10);
  putchar(a % 10 + '0');
}
template <class T>
inline void outln(T a) {
  out(a);
  puts("");
}
template <class T>
inline void out(T a, T b) {
  out(a);
  putchar(' ');
  out(b);
}
template <class T>
inline void outln(T a, T b) {
  out(a);
  putchar(' ');
  outln(b);
}
template <class T>
inline void out(T a, T b, T c) {
  out(a);
  putchar(' ');
  out(b);
  putchar(' ');
  out(c);
}
template <class T>
inline void outln(T a, T b, T c) {
  out(a);
  putchar(' ');
  outln(b);
  putchar(' ');
  outln(b);
}
template <class T>
inline void rmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
inline void rmax(T &a, const T &b) {
  if (a < b) a = b;
}
const int cnt_edge = 200010;
const int cnt_v = 100010;
int head[cnt_v], cnt_e;
struct EDGE {
  int v, next, cost;
} edge[cnt_edge];
void addedge(int u, int v, int cost = 0) {
  edge[cnt_e].v = v;
  edge[cnt_e].cost = cost;
  edge[cnt_e].next = head[u];
  head[u] = cnt_e++;
}
void init() {
  cnt_e = 0;
  memset(head, -1, sizeof(head));
}
struct data {
  int n;
  int p[15];
  data() { n = 0; }
  void add(int num) {
    int pos = lower_bound(p, p + n, num) - p;
    for (int i = n; i > pos; i--) p[i] = p[i - 1];
    p[pos] = num;
    n = min(n + 1, 10);
  }
  void print() {
    if (n != 0)
      out(n), putchar(' ');
    else
      outln(n);
    for (int i = 0; i < n; i++) {
      if (i == n - 1)
        outln(p[i]);
      else
        out(p[i]), putchar(' ');
    }
  }
} d[100010][25];
data operator+(const data &a, const data &b) {
  data c;
  int i = 0, j = 0;
  while (i < a.n && j < b.n && c.n < 10) {
    if (a.p[i] < b.p[j])
      c.p[c.n++] = a.p[i++];
    else
      c.p[c.n++] = b.p[j++];
  }
  while (i < a.n && c.n < 10) c.p[c.n++] = a.p[i++];
  while (j < b.n && c.n < 10) c.p[c.n++] = b.p[j++];
  return c;
}
int deep[100010];
int fa[100010][25];
int n, m, q;
void dfs(int u, int f, int dee) {
  fa[u][0] = f;
  deep[u] = dee;
  for (int i = 1; i <= 20; i++) {
    if (deep[u] < (1 << i)) break;
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
    d[u][i] = d[u][i - 1] + d[fa[u][i - 1]][i - 1];
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (v == f) continue;
    dfs(v, u, dee + 1);
  }
}
void lca(int x, int y, int k) {
  data ans;
  if (deep[x] < deep[y]) swap(x, y);
  int t = deep[x] - deep[y];
  for (int i = 0; i <= 20; i++)
    if (t & (1 << i)) {
      ans = ans + d[x][i];
      x = fa[x][i];
    }
  if (x == y) {
    ans = ans + d[x][0];
    ans.n = min(ans.n, k);
    ans.print();
    return;
  }
  for (int i = 20; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) {
      ans = ans + d[x][i] + d[y][i];
      x = fa[x][i];
      y = fa[y][i];
    }
  ans = ans + d[x][0] + d[y][0];
  x = fa[x][0];
  ans = ans + d[x][0];
  ans.n = min(ans.n, k);
  ans.print();
}
int main() {
  init();
  read(n, m, q);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    read(a, b);
    addedge(a, b);
    addedge(b, a);
  }
  for (int i = 1; i <= m; i++) {
    int idx;
    read(idx);
    d[idx][0].add(i);
  }
  dfs(1, 0, 0);
  while (q--) {
    int a, b, c;
    read(a, b, c);
    lca(a, b, c);
  }
}