Problem: There are n cities in Treeland connected with n - 1 bidirectional roads in such that a way that any city is reachable from any other; in other words, the graph of cities and roads is a tree. Treeland is preparing for a seasonal virus epidemic, and currently, they are trying to evaluate different infection scenarios.

In each scenario, several cities are initially infected with different virus species. Suppose that there are k_i virus species in the i-th scenario. Let us denote v_j the initial city for the virus j, and s_j the propagation speed of the virus j. The spread of the viruses happens in turns: first virus 1 spreads, followed by virus 2, and so on. After virus k_i spreads, the process starts again from virus 1.

A spread turn of virus j proceeds as follows. For each city x not infected with any virus at the start of the turn, at the end of the turn it becomes infected with virus j if and only if there is such a city y that:

  * city y was infected with virus j at the start of the turn;
  * the path between cities x and y contains at most s_j edges;
  * all cities on the path between cities x and y (excluding y) were uninfected with any virus at the start of the turn.



Once a city is infected with a virus, it stays infected indefinitely and can not be infected with any other virus. The spread stops once all cities are infected.

You need to process q independent scenarios. Each scenario is described by k_i virus species and m_i important cities. For each important city determine which the virus it will be infected by in the end.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of cities in Treeland.

The following n - 1 lines describe the roads. The i-th of these lines contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n) — indices of cities connecting by the i-th road. It is guaranteed that the given graph of cities and roads is a tree.

The next line contains a single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of infection scenarios. q scenario descriptions follow.

The description of the i-th scenario starts with a line containing two integers k_i and m_i (1 ≤ k_i, m_i ≤ n) — the number of virus species and the number of important cities in this scenario respectively. It is guaranteed that ∑_{i = 1}^ q k_i and ∑_{i = 1}^ q m_i do not exceed 2 ⋅ 10^5.

The following k_i lines describe the virus species. The j-th of these lines contains two integers v_j and s_j (1 ≤ v_j ≤ n, 1 ≤ s_j ≤ 10^6) – the initial city and the propagation speed of the virus species j. It is guaranteed that the initial cities of all virus species within a scenario are distinct.

The following line contains m_i distinct integers u_1, …, u_{m_i} (1 ≤ u_j ≤ n) — indices of important cities.

Output

Print q lines. The i-th line should contain m_i integers — indices of virus species that cities u_1, …, u_{m_i} are infected with at the end of the i-th scenario.

Example

Input


7
1 2
1 3
2 4
2 5
3 6
3 7
3
2 2
4 1
7 1
1 3
2 2
4 3
7 1
1 3
3 3
1 1
4 100
7 100
1 2 3


Output


1 2
1 1
1 1 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (typename vector<T>::const_iterator vi = v.begin(); vi != v.end(); ++vi) {
    if (vi != v.begin()) os << ", ";
    os << *vi;
  }
  os << "}";
  return os;
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  os << '(' << p.first << ", " << p.second << ')';
  return os;
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
template <typename T, bool maximum_mode = false>
struct RMQ {
  int n = 0, levels = 0;
  vector<T> values;
  vector<vector<int>> range_low;
  RMQ(const vector<T> &_values = {}) {
    if (!_values.empty()) build(_values);
  }
  static int largest_bit(int x) { return 31 - __builtin_clz(x); }
  int better_index(int a, int b) const {
    return (maximum_mode ? values[b] < values[a] : values[a] < values[b]) ? a
                                                                          : b;
  }
  void build(const vector<T> &_values) {
    values = _values;
    n = values.size();
    levels = largest_bit(n) + 1;
    range_low.resize(levels);
    for (int k = 0; k < levels; k++) range_low[k].resize(n - (1 << k) + 1);
    for (int i = 0; i < n; i++) range_low[0][i] = i;
    for (int k = 1; k < levels; k++)
      for (int i = 0; i <= n - (1 << k); i++)
        range_low[k][i] = better_index(range_low[k - 1][i],
                                       range_low[k - 1][i + (1 << (k - 1))]);
  }
  int query_index(int a, int b) const {
    assert(0 <= a && a < b && b <= n);
    int level = largest_bit(b - a);
    return better_index(range_low[level][a],
                        range_low[level][b - (1 << level)]);
  }
  T query_value(int a, int b) const { return values[query_index(a, b)]; }
};
struct LCA {
  int n = 0;
  vector<vector<int>> adj;
  vector<int> parent, depth, subtree_size;
  vector<int> euler, first_occurrence;
  vector<int> tour_start, tour_end, tour_list, postorder;
  vector<int> heavy_root;
  RMQ<int> rmq;
  LCA(int _n = 0) { init(_n); }
  LCA(const vector<vector<int>> &_adj) { init(_adj); }
  void init(int _n) {
    n = _n;
    adj.assign(n, {});
    parent.resize(n);
    depth.resize(n);
    subtree_size.resize(n);
    first_occurrence.resize(n);
    tour_start.resize(n);
    tour_end.resize(n);
    tour_list.resize(n);
    postorder.resize(n);
    heavy_root.resize(n);
  }
  void init(const vector<vector<int>> &_adj) {
    init(_adj.size());
    adj = _adj;
  }
  void add_edge(int a, int b) {
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  void dfs(int node, int par) {
    parent[node] = par;
    depth[node] = par < 0 ? 0 : depth[par] + 1;
    subtree_size[node] = 1;
    if (par >= 0)
      adj[node].erase(find(adj[node].begin(), adj[node].end(), par));
    for (int &child : adj[node]) {
      dfs(child, node);
      subtree_size[node] += subtree_size[child];
    }
    sort(adj[node].begin(), adj[node].end(),
         [&](int a, int b) { return subtree_size[a] > subtree_size[b]; });
  }
  int tour, post_tour;
  void tour_dfs(int node, bool heavy) {
    heavy_root[node] = heavy ? heavy_root[parent[node]] : node;
    first_occurrence[node] = euler.size();
    euler.push_back(node);
    tour_list[tour] = node;
    tour_start[node] = tour++;
    bool heavy_child = true;
    for (int child : adj[node]) {
      tour_dfs(child, heavy_child);
      euler.push_back(node);
      heavy_child = false;
    }
    tour_end[node] = tour;
    postorder[node] = post_tour++;
  }
  void build() {
    parent.assign(n, -1);
    for (int i = 0; i < n; i++)
      if (parent[i] < 0) dfs(i, -1);
    tour = post_tour = 0;
    for (int i = 0; i < n; i++)
      if (parent[i] < 0) {
        tour_dfs(i, false);
        euler.push_back(-1);
      }
    assert((int)euler.size() == 2 * n);
    vector<int> euler_depths;
    for (int node : euler)
      euler_depths.push_back(node < 0 ? node : depth[node]);
    rmq.build(euler_depths);
  }
  pair<int, int> find_farthest(int node, int par, int path) const {
    pair<int, int> current = {path, node};
    for (int neighbor : adj[node])
      if (neighbor != par)
        current = max(current, find_farthest(neighbor, node, path + 1));
    return current;
  }
  pair<int, pair<int, int>> get_diameter() const {
    int u = find_farthest(0, -1, 0).second;
    pair<int, int> farthest = find_farthest(u, -1, 0);
    int v = farthest.second;
    return {farthest.first, {u, v}};
  }
  int get_lca(int a, int b) const {
    a = first_occurrence[a];
    b = first_occurrence[b];
    if (a > b) swap(a, b);
    return euler[rmq.query_index(a, b + 1)];
  }
  bool is_ancestor(int a, int b) const {
    return tour_start[a] <= tour_start[b] && tour_start[b] < tour_end[a];
  }
  bool on_path(int x, int a, int b) const {
    return (is_ancestor(x, a) || is_ancestor(x, b)) &&
           is_ancestor(get_lca(a, b), x);
  }
  int get_dist(int a, int b) const {
    return depth[a] + depth[b] - 2 * depth[get_lca(a, b)];
  }
  int child_ancestor(int a, int b) const {
    assert(a != b);
    assert(is_ancestor(a, b));
    int child =
        euler[rmq.query_index(first_occurrence[a], first_occurrence[b] + 1) +
              1];
    assert(parent[child] == a);
    assert(is_ancestor(child, b));
    return child;
  }
  int get_kth_ancestor(int a, int k) const {
    while (a >= 0) {
      int root = heavy_root[a];
      if (depth[root] <= depth[a] - k) return tour_list[tour_start[a] - k];
      k -= depth[a] - depth[root] + 1;
      a = parent[root];
    }
    return a;
  }
  int get_kth_node_on_path(int a, int b, int k) const {
    int anc = get_lca(a, b);
    int first_half = depth[a] - depth[anc];
    int second_half = depth[b] - depth[anc];
    assert(0 <= k && k <= first_half + second_half);
    if (k < first_half)
      return get_kth_ancestor(a, k);
    else
      return get_kth_ancestor(b, first_half + second_half - k);
  }
  vector<pair<int, int>> compress_tree(vector<int> nodes) const {
    if (nodes.empty()) return {};
    auto &&compare_tour = [&](int a, int b) {
      return tour_start[a] < tour_start[b];
    };
    sort(nodes.begin(), nodes.end(), compare_tour);
    int k = nodes.size();
    for (int i = 0; i < k - 1; i++)
      nodes.push_back(get_lca(nodes[i], nodes[i + 1]));
    sort(nodes.begin(), nodes.end(), compare_tour);
    nodes.resize(unique(nodes.begin(), nodes.end()) - nodes.begin());
    k = nodes.size();
    vector<pair<int, int>> result = {{nodes[0], -1}};
    for (int i = 1; i < k; i++)
      result.emplace_back(nodes[i], get_lca(nodes[i], nodes[i - 1]));
    return result;
  }
};
const int INF = 1e9 + 5;
struct state {
  int node;
  pair<int, int> dist;
  state() {}
  state(int _node, pair<int, int> _dist) : node(_node), dist(_dist) {}
  bool operator<(const state &other) const { return dist > other.dist; }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N, Q;
  cin >> N;
  LCA lca(N);
  for (int i = 0; i < N - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    lca.add_edge(x, y);
  }
  lca.build();
  cin >> Q;
  vector<vector<int>> compressed_adj(N);
  vector<int> which_virus(N, -1);
  vector<pair<int, int>> dist(N, {INF, INF});
  for (int q = 0; q < Q; q++) {
    int k, m;
    cin >> k >> m;
    vector<pair<int, int>> viruses(k);
    vector<int> queries(m);
    vector<int> nodes;
    for (auto &virus : viruses) {
      cin >> virus.first >> virus.second;
      virus.first--;
      nodes.push_back(virus.first);
    }
    for (int &query : queries) {
      cin >> query;
      query--;
      nodes.push_back(query);
    }
    vector<pair<int, int>> compressed = lca.compress_tree(nodes);
    int n = compressed.size();
    for (int i = 0; i < n; i++) {
      int node = compressed[i].first;
      int parent = compressed[i].second;
      if (parent >= 0) {
        compressed_adj[node].push_back(parent);
        compressed_adj[parent].push_back(node);
      }
    }
    priority_queue<state> pq;
    auto &&dijkstra_check = [&](int node, pair<int, int> new_dist) {
      if (new_dist < dist[node]) {
        dist[node] = new_dist;
        which_virus[node] = new_dist.second;
        pq.emplace(node, new_dist);
      }
    };
    for (int i = 0; i < k; i++) {
      int node = viruses[i].first;
      dijkstra_check(node, {0, i});
    }
    while (!pq.empty()) {
      state top = pq.top();
      pq.pop();
      if (top.dist > dist[top.node]) continue;
      int virus = which_virus[top.node];
      int speed = viruses[virus].second;
      for (int neighbor : compressed_adj[top.node]) {
        int virus_dist = lca.get_dist(viruses[virus].first, neighbor);
        int time = (virus_dist + speed - 1) / speed;
        dijkstra_check(neighbor, {time, virus});
      }
    }
    for (int i = 0; i < m; i++)
      cout << which_virus[queries[i]] + 1 << (i < m - 1 ? ' ' : '\n');
    for (int i = 0; i < n; i++) {
      int node = compressed[i].first;
      compressed_adj[node].clear();
      which_virus[node] = -1;
      dist[node] = {INF, INF};
    }
  }
}