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;
const long long MOD = 1e9 + 7;
template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
struct LowestCommonAncestor {
  vector<vector<int>> g;
  int root;
  vector<int> parent[25];
  vector<int> depth;
  LowestCommonAncestor() {}
  LowestCommonAncestor(int V, int root_) : g(V), depth(V) {
    root = root_;
    for (int i = 0; i < 25; i++) parent[i].resize(V);
  }
  void initialize(int V, int root_) {
    g.resize(V);
    depth.resize(V, 0);
    root = root_;
    for (int i = 0; i < 25; i++) parent[i].resize(V);
  }
  void add_edge(int u, int v) {
    g[u].push_back(v);
    g[v].push_back(u);
  }
  void dfs(int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (int i = 0; i < g[v].size(); i++) {
      if (g[v][i] != p) dfs(g[v][i], v, d + 1);
    }
  }
  void build(int V) {
    dfs(root, -1, 0);
    for (int k = 0; k + 1 < 25; k++) {
      for (int v = 0; v < V; v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  int query(int u, int v) {
    if (depth[u] > depth[v]) swap(u, v);
    for (int k = 0; k < 25; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = 25 - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
vector<int> order;
vector<vector<int>> g;
int cnt;
void dfs(int now, int par) {
  order[now] = cnt++;
  for (auto nex : g[now])
    if (nex != par) dfs(nex, now);
}
vector<int> LCA_AuxiliaryTree(vector<int> nodes, vector<vector<int>> &g,
                              LowestCommonAncestor &lca) {
  sort(nodes.begin(), nodes.end(),
       [&](int a, int b) { return order[a] < order[b]; });
  int sz = nodes.size();
  for (int i = 0; i + 1 < sz; i++)
    nodes.push_back(lca.query(nodes[i], nodes[i + 1]));
  sort(nodes.begin(), nodes.end(),
       [&](int a, int b) { return order[a] < order[b]; });
  for (int i = 0; i < nodes.size(); i++) g[nodes[i]].clear();
  for (int i = 1; i < nodes.size(); i++) {
    int par = lca.query(nodes[i - 1], nodes[i]);
    g[par].push_back(nodes[i]);
    g[nodes[i]].push_back(par);
  }
  return nodes;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  LowestCommonAncestor lca(N, 0);
  g.resize(N);
  for (int i = 0; i < (N - 1); ++i) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    lca.add_edge(a, b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  lca.build(N);
  order.resize(N);
  cnt = 0;
  dfs(0, -1);
  vector<vector<int>> reducedTree(N);
  vector<tuple<int, int, int>> dist(N);
  int Q;
  cin >> Q;
  for (int q = 0; q < (Q); ++q) {
    int K, M;
    cin >> K >> M;
    vector<int> V(K), S(K);
    for (int i = 0; i < (K); ++i) cin >> V[i] >> S[i];
    vector<int> U(M);
    for (int i = 0; i < (M); ++i) cin >> U[i];
    for (int i = 0; i < (K); ++i) V[i]--;
    for (int i = 0; i < (M); ++i) U[i]--;
    vector<int> v(K + M);
    for (int i = 0; i < (K); ++i) v[i] = V[i];
    for (int i = 0; i < (M); ++i) v[i + K] = U[i];
    vector<int> nodes = LCA_AuxiliaryTree(v, reducedTree, lca);
    auto dijkstra = [&]() {
      priority_queue<tuple<int, int, int, int>,
                     vector<tuple<int, int, int, int>>,
                     greater<tuple<int, int, int, int>>>
          PQ;
      for (int i = 0; i < (nodes.size()); ++i)
        dist[nodes[i]] = make_tuple(MOD, MOD, MOD);
      for (int i = 0; i < (K); ++i) {
        PQ.push(make_tuple(0, i, 0, V[i]));
        dist[V[i]] = make_tuple(0, i, 0);
      }
      while (!PQ.empty()) {
        auto f = PQ.top();
        PQ.pop();
        int turn = get<0>(f);
        int virus = get<1>(f);
        int d = get<2>(f);
        int now = get<3>(f);
        if (dist[now] < make_tuple(turn, virus, d)) continue;
        if (d == 0) turn++;
        for (auto nex : reducedTree[now]) {
          int neco = lca.depth[now] + lca.depth[nex] -
                     2 * lca.depth[lca.query(now, nex)];
          if (d + neco <= S[virus]) {
            if (dist[nex] > make_tuple(turn, virus, d + neco)) {
              dist[nex] = make_tuple(turn, virus, d + neco);
              PQ.push(make_tuple(turn, virus, d + neco, nex));
            }
            continue;
          }
          int rest = neco - (S[virus] - d);
          int times = (rest - 1) / S[virus];
          int np = rest - S[virus] * times;
          if (dist[nex] > make_tuple(turn + 1 + times, virus, np)) {
            dist[nex] = make_tuple(turn + 1 + times, virus, np);
            PQ.push(make_tuple(turn + 1 + times, virus, np, nex));
          }
        }
      }
    };
    dijkstra();
    for (int i = 0; i < M; i++) cout << get<1>(dist[U[i]]) + 1 << " ";
    cout << "\n";
  }
  return 0;
}