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;
vector<int> v[100005], city[100005];
int tin[100005], tout[100005], par[100005][17];
int luu[100005][17][10];
int peo[100005][17];
int dtime, maxbit;
inline void inp(int &x) {
  x = 0;
  int c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
}
inline void out(int x) {
  if (x < 10)
    putchar(x + '0');
  else
    out(x / 10), putchar(x % 10 + '0');
}
inline void gop1(int c[], int &sz, vector<int> a, vector<int> b) {
  int res[50], cnt = 0;
  int i, j, sza = a.size(), szb = b.size();
  for (i = 0, j = 0; (i < sza || j < szb);) {
    if (i == sza)
      res[cnt++] = b[j++];
    else if (j == szb)
      res[cnt++] = a[i++];
    else if (a[i] < b[j])
      res[cnt++] = a[i++];
    else
      res[cnt++] = b[j++];
  }
  sz = unique(res, res + cnt) - res;
  sz = ((10) < (sz) ? (10) : (sz));
  for ((i) = (((10) < (sz) ? (10) : (sz)) - 1); (i) >= 0; (i)--) c[i] = res[i];
}
inline void gop(int c[], int &sz, int a[], int sza, int b[], int szb) {
  int res[50], cnt = 0;
  int i, j;
  for (i = 0, j = 0; (i < sza || j < szb);) {
    if (i == sza)
      res[cnt++] = b[j++];
    else if (j == szb)
      res[cnt++] = a[i++];
    else if (a[i] < b[j])
      res[cnt++] = a[i++];
    else
      res[cnt++] = b[j++];
  }
  sz = unique(res, res + cnt) - res;
  sz = ((10) < (sz) ? (10) : (sz));
  for ((i) = (sz - 1); (i) >= 0; (i)--) c[i] = res[i];
}
inline void dfs(int x, int g) {
  par[x][0] = g;
  gop1(luu[x][0], peo[x][0], city[x], city[g]);
  tin[x] = ++dtime;
  for (typeof(v[x].begin()) ite = v[x].begin(); ite != v[x].end(); ite++)
    if (*ite != g) dfs(*ite, x);
  tout[x] = dtime;
}
inline bool is_parent(int x, int y) {
  return tin[x] <= tin[y] && tout[y] <= tout[x];
}
inline void qr(int x, int y, int z) {
  int i;
  int res[50], lenres = 0;
  if (x == y) {
    lenres = city[x].size();
    for ((i) = 0; (i) < (lenres); (i)++) res[i] = city[x][i];
  } else if (is_parent(x, y)) {
    for ((i) = (maxbit - 1); (i) >= 0; (i)--)
      if (!is_parent(par[y][i], x))
        gop(res, lenres, res, lenres, luu[y][i], peo[y][i]), y = par[y][i];
    gop(res, lenres, res, lenres, luu[y][0], peo[y][0]);
  } else if (is_parent(y, x)) {
    for ((i) = (maxbit - 1); (i) >= 0; (i)--)
      if (!is_parent(par[x][i], y))
        gop(res, lenres, res, lenres, luu[x][i], peo[x][i]), x = par[x][i];
    gop(res, lenres, res, lenres, luu[x][0], peo[x][0]);
  } else {
    for ((i) = (maxbit - 1); (i) >= 0; (i)--)
      if (!is_parent(par[y][i], x))
        gop(res, lenres, res, lenres, luu[y][i], peo[y][i]), y = par[y][i];
    gop(res, lenres, res, lenres, luu[y][0], peo[y][0]);
    for ((i) = (maxbit - 1); (i) >= 0; (i)--)
      if (!is_parent(par[x][i], y))
        gop(res, lenres, res, lenres, luu[x][i], peo[x][i]), x = par[x][i];
    gop(res, lenres, res, lenres, luu[x][0], peo[x][0]);
  }
  lenres = ((lenres) < (z) ? (lenres) : (z));
  out(lenres);
  for ((i) = 0; (i) < (lenres); (i)++) putchar(' '), out(res[i]);
  putchar('\n');
}
inline void input() {
  int n, nm, nq, x, y, z, i, j;
  inp(n), inp(nm), inp(nq);
  for ((i) = (2); (i) <= (n); (i)++) {
    inp(x), inp(y);
    v[x].push_back(y), v[y].push_back(x);
  }
  for ((i) = (1); (i) <= (nm); (i)++) {
    inp(x);
    city[x].push_back(i);
  }
  for ((i) = (1); (i) <= (n); (i)++)
    if (city[i].size() > 10) city[i].resize(10);
  dtime = 0;
  dfs(1, 1);
  for (maxbit = 0; 1 << maxbit <= n; maxbit++)
    ;
  for (j = 1; j < maxbit; j++)
    for ((i) = (1); (i) <= (n); (i)++) {
      x = par[i][j - 1];
      par[i][j] = par[x][j - 1];
      gop(luu[i][j], peo[i][j], luu[i][j - 1], peo[i][j - 1], luu[x][j - 1],
          peo[x][j - 1]);
    }
  while (nq--) {
    inp(x), inp(y), inp(z);
    qr(x, y, z);
  }
}
int main() {
  input();
  return 0;
}