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;
struct E {
  int to, w;
};
const int N = 2e5 + 5, L = 20, INF = 1e9;
vector<int> g[N];
vector<E> compr_g[N];
int tin[N], tout[N], anc[N][L], h[N], dt[N], ind[N], rem[N], speed[N];
int timer = 0;
struct Comparator {
  bool operator()(const int &x, const int &y) {
    return dt[x] == dt[y] ? ind[x] > ind[y] : dt[x] > dt[y];
  }
};
void dfs(int node) {
  tin[node] = ++timer;
  for (int i = 1; i < L; i++) {
    anc[node][i] = anc[anc[node][i - 1]][i - 1];
  }
  for (int to : g[node]) {
    if (to != anc[node][0]) {
      h[to] = h[node] + 1;
      anc[to][0] = node;
      dfs(to);
    }
  }
  tout[node] = timer;
}
bool upper(int x, int y) { return tin[x] <= tin[y] && tout[x] >= tout[y]; }
int get_lca(int x, int y) {
  if (upper(x, y)) return x;
  for (int i = L - 1; i >= 0; i--) {
    if (!upper(anc[x][i], y)) {
      x = anc[x][i];
    }
  }
  return anc[x][0];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int k, m;
    cin >> k >> m;
    vector<int> important_cities, vertices, viruses;
    for (int j = 0; j < k; j++) {
      int x, y;
      cin >> x >> y;
      x--;
      speed[j] = y;
      viruses.push_back(x);
      vertices.push_back(x);
    }
    for (int j = 0; j < m; j++) {
      int x;
      cin >> x;
      x--;
      important_cities.push_back(x);
      vertices.push_back(x);
    }
    sort(vertices.begin(), vertices.end(),
         [&](const int &x, const int &y) { return tin[x] < tin[y]; });
    for (int i = vertices.size() - 1; i > 0; i--) {
      vertices.push_back(get_lca(vertices[i], vertices[i - 1]));
    }
    sort(vertices.begin(), vertices.end());
    vertices.erase(unique(vertices.begin(), vertices.end()), vertices.end());
    sort(vertices.begin(), vertices.end(),
         [&](const int &x, const int &y) { return tin[x] < tin[y]; });
    vector<int> stk = {vertices[0]};
    for (int j = 1; j < vertices.size(); j++) {
      while (tout[stk.back()] < tout[vertices[j]]) {
        stk.pop_back();
      }
      int distance = h[vertices[j]] - h[stk.back()];
      compr_g[stk.back()].push_back({vertices[j], distance});
      compr_g[vertices[j]].push_back({stk.back(), distance});
      stk.push_back(vertices[j]);
    }
    priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int>>,
                   greater<pair<pair<int, int>, int>>>
        q;
    for (int v : vertices) {
      dt[v] = INF;
    }
    for (int j = 0; j < viruses.size(); j++) {
      dt[viruses[j]] = 0;
      ind[viruses[j]] = j;
      rem[viruses[j]] = 0;
      q.push({{dt[viruses[j]], ind[viruses[j]]}, viruses[j]});
    }
    while (!q.empty()) {
      auto p = q.top();
      q.pop();
      if (dt[p.second] != p.first.first || ind[p.second] != p.first.second) {
        continue;
      }
      int node = p.second;
      for (E e : compr_g[node]) {
        int move_count;
        if (e.w <= rem[node]) {
          move_count = 0;
        } else {
          move_count =
              (e.w - rem[node] + speed[ind[node]] - 1) / speed[ind[node]];
        }
        int nrem = move_count * speed[ind[node]] + rem[node] - e.w,
            ndt = dt[node] + move_count;
        if (dt[e.to] > ndt || dt[e.to] == ndt && ind[e.to] > ind[node]) {
          dt[e.to] = ndt;
          ind[e.to] = ind[node];
          rem[e.to] = nrem;
          q.push({{dt[e.to], ind[e.to]}, e.to});
        }
      }
    }
    for (int v : important_cities) {
      cout << ind[v] + 1 << " ";
    }
    cout << "\n";
    for (int v : vertices) {
      compr_g[v].clear();
    }
  }
  return 0;
}