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>
int n, q, clk;
std::vector<int> sz, top, dep, parent, nodes, id, city, speed, imp, dfn, stk,
    tim, span;
std::vector<std::vector<int>> e;
std::set<std::tuple<int, int, int>> que;
void dfsSz(int u) {
  if (parent[u] != -1)
    e[u].erase(std::find(e[u].begin(), e[u].end(), parent[u]));
  sz[u] = 1;
  for (int &v : e[u]) {
    parent[v] = u;
    dep[v] = dep[u] + 1;
    dfsSz(v);
    sz[u] += sz[v];
    if (sz[v] > sz[e[u][0]]) std::swap(v, e[u][0]);
  }
}
void dfsHLD(int u) {
  dfn[u] = clk++;
  for (int v : e[u]) {
    if (v == e[u][0]) {
      top[v] = top[u];
    } else {
      top[v] = v;
    }
    dfsHLD(v);
  }
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] > dep[top[v]]) {
      u = parent[top[u]];
    } else {
      v = parent[top[v]];
    }
  }
  if (dep[u] < dep[v]) {
    return u;
  } else {
    return v;
  }
}
void solve(int u) {
  while (!e[u].empty() && id[e[u].back()] != -1) e[u].pop_back();
  if (e[u].empty()) return;
  int v = e[u].back();
  int dis = std::abs(dep[u] - dep[v]) - span[u];
  int day = tim[u] + std::max(0, dis + speed[id[u]] - 1) / speed[id[u]];
  que.emplace(day, id[u], u);
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> n;
  e.resize(n);
  sz.resize(n);
  dep.resize(n);
  top.resize(n);
  parent.resize(n);
  dfn.resize(n);
  id.resize(n);
  tim.resize(n);
  span.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    std::cin >> u >> v;
    --u;
    --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  parent[0] = -1;
  dfsSz(0);
  dfsHLD(0);
  for (int i = 0; i < n; ++i) e[i].clear();
  std::cin >> q;
  for (int i = 0; i < q; ++i) {
    int k, m;
    std::cin >> k >> m;
    nodes.clear();
    city.resize(k);
    speed.resize(k);
    for (int j = 0; j < k; ++j) {
      std::cin >> city[j] >> speed[j];
      --city[j];
      nodes.push_back(city[j]);
    }
    imp.resize(m);
    for (int j = 0; j < m; ++j) {
      std::cin >> imp[j];
      --imp[j];
      nodes.push_back(imp[j]);
    }
    nodes.push_back(0);
    std::sort(nodes.begin(), nodes.end(),
              [&](int u, int v) { return dfn[u] < dfn[v]; });
    nodes.erase(std::unique(nodes.begin(), nodes.end()), nodes.end());
    stk = {0};
    int cnt = nodes.size();
    for (int i = 1; i < cnt; ++i) {
      int u = nodes[i];
      int l = lca(stk.back(), u);
      while (int(stk.size()) >= 2 && dep[l] <= dep[stk[stk.size() - 2]]) {
        e[stk[stk.size() - 2]].push_back(stk.back());
        e[stk.back()].push_back(stk[stk.size() - 2]);
        stk.pop_back();
      }
      if (l != stk.back()) {
        nodes.push_back(l);
        e[l].push_back(stk.back());
        e[stk.back()].push_back(l);
        stk.back() = l;
      }
      stk.push_back(u);
    }
    while (int(stk.size()) >= 2) {
      int u = stk.back();
      stk.pop_back();
      e[stk.back()].push_back(u);
      e[u].push_back(stk.back());
    }
    for (int u : nodes) id[u] = -1;
    for (int u : nodes)
      std::sort(e[u].begin(), e[u].end(), [&](int i, int j) {
        return std::abs(dep[i] - dep[u]) > std::abs(dep[j] - dep[u]);
      });
    for (int j = 0; j < k; ++j) {
      id[city[j]] = j;
      tim[city[j]] = 0;
      span[city[j]] = 0;
    }
    for (int j = 0; j < k; ++j) solve(city[j]);
    while (!que.empty()) {
      auto [day, vir, u] = *que.begin();
      que.erase(que.begin());
      while (!e[u].empty()) {
        int v = e[u].back();
        if (id[v] != -1) {
          e[u].pop_back();
          continue;
        }
        int dis = std::abs(dep[u] - dep[v]);
        if (span[u] + (day - tim[u]) * speed[id[u]] < dis) break;
        e[u].pop_back();
        id[v] = id[u];
        tim[v] = day;
        span[v] = span[u] + (day - tim[u]) * speed[id[u]] - dis;
        solve(v);
      }
      solve(u);
    }
    for (int j = 0; j < m; ++j)
      std::cout << id[imp[j]] + 1 << " \n"[j == m - 1];
  }
  return 0;
}