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>
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
int dep[maxn], f[maxn][32], cnt, T[maxn];
int n, m, q;
vector<int> num[maxn], v[maxn];
vector<int> ans;
struct node {
  int sum, l, r;
  node() { sum = l = r = 0; }
} t[maxn << 6];
void update(int pre, int &now, int l, int r, int pos) {
  now = ++cnt;
  t[now].sum = t[pre].sum + 1;
  if (l == r) return;
  t[now].l = t[pre].l;
  t[now].r = t[pre].r;
  int mid = l + r >> 1;
  if (mid >= pos)
    update(t[pre].l, t[now].l, l, mid, pos);
  else
    update(t[pre].r, t[now].r, mid + 1, r, pos);
}
void dfs(int x, int p, int d) {
  dep[x] = d;
  f[x][0] = p;
  for (int i = (1); i < (30); i++) f[x][i] = f[f[x][i - 1]][i - 1];
  T[x] = T[p];
  for (auto u : num[x]) update(T[x], T[x], 1, m, u);
  for (auto u : v[x])
    if (u != p) dfs(u, x, d + 1);
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 29, d = dep[u] - dep[v]; i >= 0; i--)
    if (d >> i & 1) u = f[u][i];
  if (u == v) return v;
  for (int i = 29; i >= 0; i--)
    if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
  return f[u][0];
}
void qr(int x, int y, int z, int d, int l, int r, int k) {
  int sl = t[t[x].l].sum + t[t[y].l].sum - t[t[z].l].sum - t[t[d].l].sum;
  if (l == r) {
    if (k) ans.push_back(l);
    return;
  }
  int mid = l + r >> 1;
  if (sl >= k)
    qr(t[x].l, t[y].l, t[z].l, t[d].l, l, mid, k);
  else {
    if (sl) qr(t[x].l, t[y].l, t[z].l, t[d].l, l, mid, sl);
    qr(t[x].r, t[y].r, t[z].r, t[d].r, mid + 1, r, k - sl);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = (0); i < (n - 1); i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = (0); i < (m); i++) {
    int x;
    cin >> x;
    num[x].push_back(i + 1);
  }
  dfs(1, 1, 0);
  for (int i = (0); i < (q); i++) {
    int x, y, a;
    ans.clear();
    cin >> x >> y >> a;
    int l = lca(x, y), d = f[l][0];
    if (l == 1) d = 0;
    int k = min(a, t[T[x]].sum + t[T[y]].sum - t[T[l]].sum - t[T[d]].sum);
    qr(T[x], T[y], T[l], T[d], 1, m, k);
    cout << (int)ans.size();
    for (auto i : ans) cout << " " << i;
    cout << endl;
  }
  return 0;
}
