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 int L = 18, N_ = 200100;
int N, Q, Num, Par[N_][L + 1], Dep[N_], St[N_], End[N_], Nds[N_ * 2], All[N_],
    Tulos[N_], Vis[N_];
vector<int> V[N_];
vector<pair<int, int>> E[200100];
int LCA(int a, int b) {
  if (Dep[b] > Dep[a]) return LCA(b, a);
  for (int i = L; i >= 0; --i)
    if (Dep[Par[a][i]] >= Dep[b]) a = Par[a][i];
  for (int i = L; i >= 0; --i)
    if (Par[a][i] != Par[b][i]) a = Par[a][i], b = Par[b][i];
  if (a != b) return Par[a][0];
  return a;
}
void Haku(int s, int e) {
  Dep[s] = Dep[e] + 1;
  Par[s][0] = e;
  St[s] = ++Num;
  for (int i = 1; i <= L; ++i) Par[s][i] = Par[Par[s][i - 1]][i - 1];
  for (auto &u : V[s])
    if (u != e) Haku(u, s);
  End[s] = Num;
}
int InSubtree(int a, int b) { return St[a] <= St[b] && St[b] <= End[a]; }
void CreateInduced(int K) {
  int i, j;
  sort(Nds + 1, Nds + 1 + K, [&](int a, int b) { return St[a] < St[b]; });
  K = unique(Nds + 1, Nds + 1 + K) - Nds - 1;
  j = 0;
  for (i = 1; i <= K; ++i) {
    All[++j] = Nds[i];
    if (i + 1 > K) break;
    All[++j] = LCA(Nds[i], Nds[i + 1]);
  }
  K = j;
  sort(All + 1, All + 1 + K, [&](int a, int b) { return St[a] < St[b]; });
  K = unique(All + 1, All + 1 + K) - All - 1;
  vector<int> St{All[1]};
  for (i = 1; i <= K; ++i) {
    E[All[i]].clear();
    Tulos[All[i]] = 0;
    Vis[All[i]] = 0;
  }
  for (i = 2; i <= K; ++i) {
    while (St.size() && !InSubtree(St.back(), All[i])) St.pop_back();
    E[St.back()].emplace_back(All[i], Dep[All[i]] - Dep[St.back()]);
    E[All[i]].emplace_back(St.back(), Dep[All[i]] - Dep[St.back()]);
    St.push_back(All[i]);
  }
}
struct State {
  int Round, City, Spread, Origin;
  bool operator<(const State &x) const {
    if (Round != x.Round) return Round > x.Round;
    if (Origin != x.Origin) return Origin > x.Origin;
    return City > x.City;
  }
};
void CalcCities(vector<pair<int, int>> &Inf) {
  int i;
  priority_queue<State> pq;
  for (i = 0; i < (int)Inf.size(); ++i) pq.push({0, Inf[i].first, 0, i});
  for (; !pq.empty();) {
    auto a = pq.top();
    pq.pop();
    if (Vis[a.City]) continue;
    Vis[a.City] = 1;
    Tulos[a.City] = a.Origin + 1;
    for (auto &[u, w] : E[a.City])
      if (!Vis[u]) {
        if (a.Spread < w) {
          int U =
              (w - a.Spread + Inf[a.Origin].second - 1) / Inf[a.Origin].second;
          pq.push({a.Round + U, u, U * Inf[a.Origin].second - w + a.Spread,
                   a.Origin});
        } else
          pq.push({a.Round, u, a.Spread - w, a.Origin});
      }
  }
}
void Solve() {
  int i;
  int A, B;
  scanf("%d%d", &A, &B);
  int it = 0;
  vector<pair<int, int>> Inf;
  for (i = 1; i <= A; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    Inf.emplace_back(a, b);
    Nds[++it] = a;
  }
  vector<int> Imp;
  for (i = 1; i <= B; ++i) {
    int x;
    scanf("%d", &x);
    Imp.push_back(x);
    Nds[++it] = x;
  }
  CreateInduced(it);
  CalcCities(Inf);
  for (i = 0; i < B; ++i) printf("%d ", Tulos[Imp[i]]);
  puts("");
}
int main() {
  int i;
  scanf("%d", &N);
  for (i = 2; i <= N; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  Haku(1, 1);
  scanf("%d", &Q);
  while (Q--) Solve();
}