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;
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, ms63 = 1061109567;
int casenum, casei;
int n, m, q, tim;
int x, y, w;
struct A {
  int l, r;
  int g;
  int b[15];
} c[1 << 18];
vector<int> a[N];
int g[N];
int b[N][10];
int G, tmpg;
int B[10], tmpb[10];
int fa[N], son[N], dep[N], siz[N], pos[N], rev[N], top[N];
void dfs1(int x) {
  siz[x] = 1;
  son[x] = 0;
  for (int i = a[x].size() - 1; i >= 0; i--) {
    int y = a[x][i];
    if (y == fa[x]) continue;
    fa[y] = x;
    dep[y] = dep[x] + 1;
    dfs1(y);
    siz[x] += siz[y];
    if (siz[y] > siz[son[x]]) son[x] = y;
  }
}
void dfs2(int x, int chain) {
  pos[x] = ++tim;
  rev[tim] = x;
  top[x] = chain;
  if (son[x] == 0) return;
  dfs2(son[x], chain);
  for (int i = a[x].size() - 1; i >= 0; i--) {
    y = a[x][i];
    if (y != fa[x] && y != son[x]) dfs2(y, y);
  }
}
void merg(int &g, int &gl, int &gr, int b[], int bl[], int br[]) {
  int p = 0, pl = 0, pr = 0;
  while (p < 10) {
    if (pl < gl && pr < gr) {
      if (bl[pl] < br[pr])
        b[p++] = bl[pl++];
      else
        b[p++] = br[pr++];
    } else if (pl < gl)
      b[p++] = bl[pl++];
    else if (pr < gr)
      b[p++] = br[pr++];
    else
      break;
  }
  g = p;
}
void pushup(int o) {
  merg(c[o].g, c[o << 1].g, c[o << 1 | 1].g, c[o].b, c[o << 1].b,
       c[o << 1 | 1].b);
}
void build(int o, int l, int r) {
  c[o].l = l;
  c[o].r = r;
  if (l == r) {
    int v = rev[l];
    c[o].g = g[v];
    memcpy(c[o].b, b[v], sizeof(c[o].b));
    return;
  }
  int m = (l + r) / 2;
  build(o << 1, l, m);
  build(o << 1 | 1, m + 1, r);
  pushup(o);
}
void update(int o, int l, int r) {
  if (c[o].l == l && c[o].r == r) {
    tmpg = G;
    memcpy(tmpb, B, sizeof(tmpb));
    merg(G, tmpg, c[o].g, B, tmpb, c[o].b);
    return;
  }
  int m = (c[o].l + c[o].r) / 2;
  if (r <= m)
    update(o << 1, l, r);
  else if (l > m)
    update(o << 1 | 1, l, r);
  else {
    update(o << 1, l, m);
    update(o << 1 | 1, m + 1, r);
  }
}
void SOLVE(int x, int y, int w) {
  G = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    update(1, pos[top[x]], pos[x]);
    x = fa[top[x]];
  }
  if (pos[x] > pos[y]) swap(x, y);
  update(1, pos[x], pos[y]);
  G = min(G, w);
  printf("%d", G);
  for (int i = 0; i < G; i++) {
    printf(" %d", B[i]);
  }
  printf("\n");
}
int main() {
  while (~scanf("%d%d%d", &n, &m, &q)) {
    for (int i = 1; i <= n; i++) {
      a[i].clear();
      g[i] = 0;
    }
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &x, &y);
      a[x].push_back(y);
      a[y].push_back(x);
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d", &x);
      if (g[x] < 10) b[x][g[x]++] = i;
    }
    dep[1] = 0;
    fa[1] = 0;
    dfs1(1);
    tim = 0;
    dfs2(1, 1);
    build(1, 1, n);
    for (int i = 1; i <= q; i++) {
      scanf("%d%d%d", &x, &y, &w);
      SOLVE(x, y, w);
    }
  }
}