#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int read() {
  char c;
  int ans = 0, f = 1;
  c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
struct edge {
  int v, w, next;
} e[100005 << 1];
int tot, head[100005];
int dep[100005], pre[100005], siz[100005], son[100005], top[100005], len,
    w[300005];
int id[100005], K, M, n;
int vt[100005][11];
int q[300005][11];
void addedge(int u, int v) {
  e[tot].v = v;
  e[tot].next = head[u];
  head[u] = tot++;
}
void dfs(int u, int fa, int d) {
  int v;
  siz[u] = 1;
  pre[u] = fa;
  dep[u] = d;
  for (int i = head[u]; ~i; i = e[i].next) {
    v = e[i].v;
    if (v == fa) continue;
    dfs(v, u, d + 1);
    siz[u] += siz[v];
    if (son[u] == -1 || siz[v] > siz[son[u]]) son[u] = v;
  }
}
void DFS(int u, int tp) {
  top[u] = tp;
  if (tp != u || son[u] != -1 || u == 1) {
    id[u] = ++len;
    w[len] = u;
  }
  if (son[u] == -1) return;
  DFS(son[u], tp);
  int v;
  for (int i = head[u]; ~i; i = e[i].next) {
    v = e[i].v;
    if (v != son[u] && v != pre[u]) DFS(v, v);
  }
}
void add(int a[], int b[], int k) {
  int ln = min(a[0] + b[0], k);
  int l1 = 1, l2 = 1;
  int c[11];
  c[0] = 0;
  while (ln--) {
    if (l1 > a[0])
      c[++c[0]] = b[l2++];
    else if (l2 > b[0])
      c[++c[0]] = a[l1++];
    else if (a[l1] < b[l2])
      c[++c[0]] = a[l1++];
    else
      c[++c[0]] = b[l2++];
  }
  for (int i = 0; i <= c[0]; i++) a[i] = c[i];
}
void build() {
  for (int i = 1; i <= len; i++) {
    q[i + M][0] = vt[w[i]][0];
    for (int j = 1; j <= vt[w[i]][0]; j++) q[i + M][j] = vt[w[i]][j];
  }
  for (int u = M - 1; u > 0; u--) {
    add(q[u], q[u << 1], 10);
    add(q[u], q[(u << 1) | 1], 10);
  }
}
int sum[11];
void query(int s, int t, int k) {
  for (s = s + M - 1, t = t + M + 1; s ^ t ^ 1; s >>= 1, t >>= 1) {
    if (~s & 1) add(sum, q[s ^ 1], k);
    if (t & 1) add(sum, q[t ^ 1], k);
  }
}
void answer(int x, int y, int k) {
  sum[0] = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    if (top[x] == x) {
      add(sum, vt[x], k);
    } else {
      query(id[top[x]], id[x], k);
    }
    x = pre[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  if (y == x) {
    add(sum, vt[x], k);
  } else {
    query(id[x], id[y], k);
  }
  int ans = min(sum[0], k);
  printf("%d", ans);
  for (int i = 1; i <= ans; i++) printf(" %d", sum[i]);
  printf("\n");
}
void init() {
  tot = 0;
  len = 0;
  memset((head), (-1), sizeof((head)));
  memset((son), (-1), sizeof((son)));
}
int main() {
  n = read();
  int m = read();
  K = read();
  int u, v, w;
  init();
  for (int i = 1; i < n; i++) {
    u = read();
    v = read();
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 0, 0);
  DFS(1, 1);
  for (int i = 1; i <= m; i++) {
    int x = read();
    if (vt[x][0] < 10) vt[x][++vt[x][0]] = i;
  }
  for (M = 1; M <= len + 1; M <<= 1)
    ;
  build();
  while (K--) {
    u = read();
    v = read();
    w = read();
    answer(u, v, w);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int read() {
  char c;
  int ans = 0, f = 1;
  c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    ans = ans * 10 + c - '0';
    c = getchar();
  }
  return ans * f;
}
struct edge {
  int v, w, next;
} e[100005 << 1];
int tot, head[100005];
int dep[100005], pre[100005], siz[100005], son[100005], top[100005], len,
    w[300005];
int id[100005], K, M, n;
int vt[100005][11];
int q[300005][11];
void addedge(int u, int v) {
  e[tot].v = v;
  e[tot].next = head[u];
  head[u] = tot++;
}
void dfs(int u, int fa, int d) {
  int v;
  siz[u] = 1;
  pre[u] = fa;
  dep[u] = d;
  for (int i = head[u]; ~i; i = e[i].next) {
    v = e[i].v;
    if (v == fa) continue;
    dfs(v, u, d + 1);
    siz[u] += siz[v];
    if (son[u] == -1 || siz[v] > siz[son[u]]) son[u] = v;
  }
}
void DFS(int u, int tp) {
  top[u] = tp;
  if (tp != u || son[u] != -1 || u == 1) {
    id[u] = ++len;
    w[len] = u;
  }
  if (son[u] == -1) return;
  DFS(son[u], tp);
  int v;
  for (int i = head[u]; ~i; i = e[i].next) {
    v = e[i].v;
    if (v != son[u] && v != pre[u]) DFS(v, v);
  }
}
void add(int a[], int b[], int k) {
  int ln = min(a[0] + b[0], k);
  int l1 = 1, l2 = 1;
  int c[11];
  c[0] = 0;
  while (ln--) {
    if (l1 > a[0])
      c[++c[0]] = b[l2++];
    else if (l2 > b[0])
      c[++c[0]] = a[l1++];
    else if (a[l1] < b[l2])
      c[++c[0]] = a[l1++];
    else
      c[++c[0]] = b[l2++];
  }
  for (int i = 0; i <= c[0]; i++) a[i] = c[i];
}
void build() {
  for (int i = 1; i <= len; i++) {
    q[i + M][0] = vt[w[i]][0];
    for (int j = 1; j <= vt[w[i]][0]; j++) q[i + M][j] = vt[w[i]][j];
  }
  for (int u = M - 1; u > 0; u--) {
    add(q[u], q[u << 1], 10);
    add(q[u], q[(u << 1) | 1], 10);
  }
}
int sum[11];
void query(int s, int t, int k) {
  for (s = s + M - 1, t = t + M + 1; s ^ t ^ 1; s >>= 1, t >>= 1) {
    if (~s & 1) add(sum, q[s ^ 1], k);
    if (t & 1) add(sum, q[t ^ 1], k);
  }
}
void answer(int x, int y, int k) {
  sum[0] = 0;
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    if (top[x] == x) {
      add(sum, vt[x], k);
    } else {
      query(id[top[x]], id[x], k);
    }
    x = pre[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  if (y == x) {
    add(sum, vt[x], k);
  } else {
    query(id[x], id[y], k);
  }
  int ans = min(sum[0], k);
  printf("%d", ans);
  for (int i = 1; i <= ans; i++) printf(" %d", sum[i]);
  printf("\n");
}
void init() {
  tot = 0;
  len = 0;
  memset((head), (-1), sizeof((head)));
  memset((son), (-1), sizeof((son)));
}
int main() {
  n = read();
  int m = read();
  K = read();
  int u, v, w;
  init();
  for (int i = 1; i < n; i++) {
    u = read();
    v = read();
    addedge(u, v);
    addedge(v, u);
  }
  dfs(1, 0, 0);
  DFS(1, 1);
  for (int i = 1; i <= m; i++) {
    int x = read();
    if (vt[x][0] < 10) vt[x][++vt[x][0]] = i;
  }
  for (M = 1; M <= len + 1; M <<= 1)
    ;
  build();
  while (K--) {
    u = read();
    v = read();
    w = read();
    answer(u, v, w);
  }
  return 0;
}

```