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> adj[100010];
int dp[100010][20];
vector<int> dp2[100010][20];
int l[100010];
vector<int> v[100010];
int pai[100010];
int n, m, q, x, y;
int a, b, c;
void ins(vector<int> &v, int x) {
  for (int i = 0; i < v.size(); ++i) {
    if (v[i] > x) {
      int aux = v[i];
      v[i] = x;
      x = aux;
      for (int j = i + 1; j < v.size(); ++j) {
        aux = v[j];
        v[j] = x;
        x = aux;
      }
      if (v.size() < 10) v.push_back(x);
      return;
    } else if (v[i] == x) {
      return;
    }
  }
  if (v.size() < 10) {
    v.push_back(x);
  }
}
void ins(vector<int> &v, vector<int> &a) {
  for (int i = 0; i < min(11, (int)a.size()); ++i) {
    ins(v, a[i]);
  }
}
void build() {
  for (int i = 0; i < n; ++i) {
    dp[i][0] = pai[i];
    ins(dp2[i][0], v[pai[i]]);
    ins(dp2[i][0], v[i]);
  }
  for (int j = 1; j < 20; ++j) {
    for (int i = 1; i < n; ++i) {
      dp[i][j] = dp[dp[i][j - 1]][j - 1];
      ins(dp2[i][j], dp2[i][j - 1]);
      ins(dp2[i][j], dp2[dp[i][j - 1]][j - 1]);
    }
  }
}
vector<int> lca(int x, int y) {
  vector<int> ret;
  ins(ret, v[x]);
  ins(ret, v[y]);
  if (l[y] > l[x]) swap(x, y);
  for (int i = 19; i >= 0; --i) {
    if (l[dp[x][i]] >= l[y]) {
      ins(ret, dp2[x][i]);
      x = dp[x][i];
    }
  }
  if (x == y) {
    ins(ret, v[x]);
    return ret;
  }
  for (int i = 19; i >= 0; --i) {
    if (dp[x][i] != dp[y][i]) {
      ins(ret, dp2[x][i]);
      ins(ret, dp2[y][i]);
      x = dp[x][i];
      y = dp[y][i];
    }
  }
  ins(ret, v[dp[x][0]]);
  return ret;
}
void dfs(int curr, int p, int len) {
  pai[curr] = p;
  l[curr] = len;
  for (int i = 0; i < adj[curr].size(); ++i) {
    int u = adj[curr][i];
    if (u != p) {
      dfs(u, curr, len + 1);
    }
  }
}
int main(void) {
  ios ::sync_with_stdio(false);
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    cin >> x;
    x--;
    v[x].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    sort(v[i].begin(), v[i].end());
  }
  dfs(0, 0, 0);
  build();
  while (q--) {
    cin >> a >> b >> c;
    a--;
    b--;
    vector<int> ans = lca(a, b);
    cout << min(c, (int)ans.size());
    for (int i = 0; i < min(c, (int)ans.size()); ++i) {
      cout << " " << ans[i];
    }
    cout << "\n";
  }
  return 0;
}