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;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int N;
  cin >> N;
  vector<vector<int>> adj(N);
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<int> depth(N);
  vector<int> st(N);
  vector<pair<int, int>> inOrder;
  inOrder.reserve(2 * N - 1);
  function<void(int, int)> dfs = [&](int cur, int prv) {
    depth[cur] = (prv == -1 ? 0 : depth[prv]) + 1;
    st[cur] = int(inOrder.size());
    for (int nxt : adj[cur]) {
      if (nxt == prv) continue;
      inOrder.emplace_back(depth[cur], cur);
      dfs(nxt, cur);
    }
    inOrder.emplace_back(depth[cur], cur);
  };
  dfs(0, -1);
  assert(int(inOrder.size()) == 2 * N - 1);
  const int S = 1 << 19;
  vector<pair<int, int>> seg(2 * S);
  for (int i = 0; i < int(inOrder.size()); i++) {
    seg[i + S] = inOrder[i];
  }
  for (int i = S - 1; i; i--) {
    seg[i] = min(seg[2 * i], seg[2 * i + 1]);
  }
  assert(0 == st[0]);
  auto get_lca = [&](int l, int r) -> int {
    if (l > r) swap(l, r);
    pair<int, int> res = inOrder[l];
    for (int a = l + S, b = r + 1 + S; a < b; a /= 2, b /= 2) {
      if (a & 1) res = min(res, seg[a++]);
      if (b & 1) res = min(res, seg[--b]);
    }
    return st[res.second];
  };
  auto get_distance = [&](int l, int r) -> int {
    int lca = get_lca(l, r);
    return inOrder[l].first + inOrder[r].first - 2 * inOrder[lca].first;
  };
  int Q;
  cin >> Q;
  while (Q--) {
    int K, M;
    cin >> K >> M;
    vector<pair<int, int>> viruses(K);
    for (auto& it : viruses) {
      cin >> it.first >> it.second;
      it.first--;
    }
    vector<int> queries(M);
    for (auto& it : queries) {
      cin >> it;
      it--;
    }
    vector<int> allCities;
    allCities.reserve(2 * (K + M));
    allCities.push_back(0);
    for (auto it : viruses) allCities.push_back(st[it.first]);
    for (auto it : queries) allCities.push_back(st[it]);
    sort(allCities.begin(), allCities.end());
    allCities.erase(unique(allCities.begin(), allCities.end()),
                    allCities.end());
    for (int i = int(allCities.size()) - 1; i > 0; i--) {
      int l = allCities[i - 1];
      int r = allCities[i];
      assert(l < r);
      int lca = get_lca(l, r);
      allCities.push_back(lca);
    }
    sort(allCities.begin(), allCities.end());
    allCities.erase(unique(allCities.begin(), allCities.end()),
                    allCities.end());
    int C = int(allCities.size());
    auto compressId = [&](int i) -> int {
      auto it = lower_bound(allCities.begin(), allCities.end(), i);
      assert(it != allCities.end() && *it == i);
      return int(it - allCities.begin());
    };
    for (auto& it : viruses) {
      it.first = compressId(st[it.first]);
    }
    for (auto& it : queries) {
      it = compressId(st[it]);
    }
    vector<vector<int>> curAdj(C);
    for (int i = 0; i + 1 < int(allCities.size()); i++) {
      int l = allCities[i];
      int r = allCities[i + 1];
      assert(l < r);
      int lca = compressId(get_lca(l, r));
      curAdj[lca].push_back(i + 1);
      curAdj[i + 1].push_back(lca);
    }
    using distVal = pair<pair<int, int>, int>;
    priority_queue<distVal, vector<distVal>, greater<distVal>> pq;
    vector<pair<int, int>> dist(C, {N + 1, -1});
    for (int i = 0; i < int(viruses.size()); i++) {
      int v = viruses[i].first;
      dist[v] = {0, i};
      pq.emplace(dist[v], v);
    }
    while (!pq.empty()) {
      auto pqD = pq.top().first;
      int cur = pq.top().second;
      pq.pop();
      assert(pqD >= dist[cur]);
      if (pqD != dist[cur]) continue;
      for (int nxt : curAdj[cur]) {
        pair<int, int> nd = {(get_distance(allCities[viruses[pqD.second].first],
                                           allCities[nxt]) +
                              viruses[pqD.second].second - 1) /
                                 viruses[pqD.second].second,
                             pqD.second};
        if (nd < dist[nxt]) {
          dist[nxt] = nd;
          pq.emplace(nd, nxt);
        }
      }
    }
    for (int i = 0; i < int(queries.size()); i++) {
      int v = queries[i];
      cout << dist[v].second + 1 << " \n"[i + 1 == int(queries.size())];
    }
  }
  return 0;
}