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>
int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  int x;
  int y, z;
  pa3(int x = 0, int y = 0, int z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  int x;
  int y, z, w;
  pa4(int x = 0, int y = 0, int z = 0, int w = 0) : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  int x, y;
  pa2(int x = 0, int y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(int i) {
  ostringstream s;
  s << i;
  return s.str();
}
int gcd(int v, int b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
double distans(double x1, double y1, double x2, double y2) {
  double rr = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
  return sqrt(rr);
}
int mod;
int extgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
int pr[1000010];
int inv[1000010];
int beki(int wa, int rr, int warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
double bekid(double w, int r) {
  if (r == 0) return 1.0;
  if (r == 1) return w;
  if (r % 2) return bekid(w, r - 1) * w;
  double f = bekid(w, r / 2);
  return f * f;
}
int comb(int nn, int rr) {
  int r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(int ert) {
  pr[0] = 1;
  for (int i = 1; i < ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  for (int i = 0; i < ert; i++) inv[i] = beki(pr[i], mod - 2, mod);
}
vector<int> G[100020], G2[100020];
int pos[100020] = {};
int dep[100020] = {};
int dp[100020][18];
void fdfs(int r, int p, int d) {
  dep[r] = d;
  for (auto v : G[r])
    if (v != p) {
      dp[v][0] = r;
      G2[r].push_back(v);
      fdfs(v, r, d + 1);
    }
}
pa4 lca(int a, int b) {
  int aa = a, bb = b;
  bool sw = 0;
  if (dep[a] < dep[b]) swap(a, b), sw = 1;
  int sa = dep[a] - dep[b];
  for (int i = 0; i < 18; i++)
    if (sa & (1 << i)) a = dp[a][i];
  if (a == b) {
    if (sw == 0)
      return {aa, a, -1, -1};
    else
      return {bb, a, -1, -1};
  }
  for (int i = 17; i >= 0; i--)
    if (dp[a][i] != dp[b][i]) {
      a = dp[a][i];
      b = dp[b][i];
    }
  if (sw == 0) {
    return {aa, a, bb, dp[b][0]};
  }
  return {aa, b, bb, dp[a][0]};
}
using A = array<int, 10>;
A merge(A a1, A a2) {
  A c;
  int i1 = 0, i2 = 0, ima = 0;
  while (1) {
    if (i1 == 10) {
      c[ima] = a2[i2];
      i2++;
    } else if (i2 == 10) {
      c[ima] = a1[i1];
      i1++;
    } else if (a1[i1] < a2[i2]) {
      c[ima] = a1[i1];
      i1++;
    } else if (a1[i1] > a2[i2]) {
      c[ima] = a2[i2];
      i2++;
    } else if (a1[i1] == 1000000007) {
      for (int i = ima; i < 10; i++) c[i] = 1000000007;
      return c;
    } else {
      c[ima] = a1[i1];
      i1++;
      i2++;
    }
    ima++;
    if (ima == 10) return c;
  }
}
A uni = {1000000007, 1000000007, 1000000007, 1000000007, 1000000007,
         1000000007, 1000000007, 1000000007, 1000000007, 1000000007};
A st[100020][18];
int n;
vector<int> doko;
void shoki2() {
  doko.push_back(0 - 1);
  for (int d = 0; d < 18; d++) {
    for (int i = 0; i < (1 << d); i++) {
      doko.push_back(d);
    }
  }
  for (int i = 1; i < 18; i++)
    for (int j = 1; j <= n; j++) {
      st[j][i] = merge(st[j][i - 1], st[dp[j][i - 1]][i - 1]);
    }
}
A kukan(int sita, int ue) {
  if (sita == ue) return st[sita][0];
  int sa = -dep[ue] + dep[sita] + 1;
  int d = doko[sa];
  int ato = sa - (1 << d);
  int e = sita;
  for (int i = 17; i >= 0; i--)
    if (ato & (1 << i)) e = dp[e][i];
  return merge(st[sita][d], st[e][d]);
}
A solve(int u, int v) {
  if (u == v) return st[u][0];
  pa4 z = lca(u, v);
  if (z.w == -1) return kukan(z.x, z.y);
  return merge(kukan(z.x, z.y), kukan(z.z, z.w));
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int m, q;
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    int y, yy;
    cin >> y >> yy;
    G[y].push_back(yy);
    G[yy].push_back(y);
  }
  fdfs(1, 1, 0);
  dp[1][0] = 1;
  for (int d = 0; d < 17; d++)
    for (int i = 1; i <= n; i++) {
      dp[i][d + 1] = dp[dp[i][d]][d];
    }
  for (int i = 0; i < n; i++) st[i + 1][0] = uni;
  for (int i = 1; i <= m; i++) {
    int y;
    cin >> y;
    if (pos[y] >= 10) continue;
    st[y][0][pos[y]] = i;
    pos[y]++;
  }
  shoki2();
  for (int er = 0; er < q; er++) {
    int u, v, a;
    cin >> u >> v >> a;
    A w = solve(u, v);
    vector<int> ve;
    for (int i = 0; i < a; i++)
      if (w[i] != 1000000007) ve.push_back(w[i]);
    cout << ve.size();
    for (auto v : ve) cout << " " << v;
    cout << "\n";
  }
  return 0;
}