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;
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
          dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b ? (a = b, true) : false;
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b ? (a = b, true) : false;
}
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;
using CostType = int;
struct Edge {
  int src, dst;
  CostType cost;
  Edge(int src, int dst, CostType cost = 0) : src(src), dst(dst), cost(cost) {}
  inline bool operator<(const Edge &x) const {
    return cost != x.cost ? cost < x.cost
           : dst != x.dst ? dst < x.dst
                          : src < x.src;
  }
  inline bool operator<=(const Edge &x) const { return !(x < *this); }
  inline bool operator>(const Edge &x) const { return x < *this; }
  inline bool operator>=(const Edge &x) const { return !(*this < x); }
};
struct LCADoubling {
  vector<int> depth;
  vector<CostType> dist;
  LCADoubling(const vector<vector<Edge>> &graph) : graph(graph) {
    n = graph.size();
    depth.resize(n);
    dist.resize(n);
    while ((1 << table_h) <= n) ++table_h;
    parent.resize(table_h, vector<int>(n));
  }
  void build(int root = 0) {
    dfs(-1, root, 0, 0);
    for (int i = 0; i + 1 < table_h; ++i)
      for (int ver = (0); ver < (n); ++ver) {
        parent[i + 1][ver] =
            (parent[i][ver] == -1 ? -1 : parent[i][parent[i][ver]]);
      }
  }
  int query(int u, int v) {
    if (depth[u] > depth[v]) swap(u, v);
    for (int i = (0); i < (table_h); ++i) {
      if ((depth[v] - depth[u]) >> i & 1) v = parent[i][v];
    }
    if (u == v) return u;
    for (int i = table_h - 1; i >= 0; --i) {
      if (parent[i][u] != parent[i][v]) {
        u = parent[i][u];
        v = parent[i][v];
      }
    }
    return parent[0][u];
  }
  CostType distance(int u, int v) {
    return dist[u] + dist[v] - dist[query(u, v)] * 2;
  }

 private:
  int n, table_h = 1;
  vector<vector<Edge>> graph;
  vector<vector<int>> parent;
  void dfs(int par, int ver, int now_depth, CostType now_dist) {
    depth[ver] = now_depth;
    dist[ver] = now_dist;
    parent[0][ver] = par;
    for (const Edge &e : graph[ver]) {
      if (e.dst != par) dfs(ver, e.dst, now_depth + 1, now_dist + e.cost);
    }
  }
};
int main() {
  int n;
  cin >> n;
  vector<vector<Edge>> graph(n);
  for (int _ = (0); _ < (n - 1); ++_) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    graph[x].emplace_back(x, y, 1);
    graph[y].emplace_back(y, x, 1);
  }
  LCADoubling lca(graph);
  lca.build(0);
  vector<int> tra(n, -1);
  function<void(int, int &)> dfs = [&](int ver, int &t) {
    tra[ver] = t++;
    for (const Edge &e : graph[ver]) {
      if (tra[e.dst] == -1) dfs(e.dst, t);
    }
  };
  int t = 0;
  dfs(0, t);
  int q;
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> v(k), s(k), u(m), vers(k + m);
    for (int i = (0); i < (k); ++i) {
      cin >> v[i] >> s[i];
      --v[i];
      vers[i] = v[i];
    }
    for (int i = (0); i < (m); ++i) {
      cin >> u[i];
      --u[i];
      vers[k + i] = u[i];
    }
    sort((vers).begin(), (vers).end(),
         [&](int a, int b) { return tra[a] < tra[b]; });
    vers.erase(unique((vers).begin(), (vers).end()), vers.end());
    for (int i = 1, sz = vers.size(); i < sz; ++i)
      vers.emplace_back(lca.query(vers[i - 1], vers[i]));
    sort((vers).begin(), (vers).end(),
         [&](int a, int b) { return tra[a] < tra[b]; });
    vers.erase(unique((vers).begin(), (vers).end()), vers.end());
    int x = vers.size();
    map<int, int> mp;
    for (int i = (0); i < (x); ++i) mp[vers[i]] = i;
    vector<vector<Edge>> g(x);
    vector<int> sta{0};
    for (int i = (1); i < (x); ++i) {
      while (lca.query(vers[sta.back()], vers[i]) != vers[sta.back()])
        sta.pop_back();
      g[sta.back()].emplace_back(sta.back(), i,
                                 lca.distance(vers[sta.back()], vers[i]));
      g[i].emplace_back(i, sta.back(), lca.distance(vers[sta.back()], vers[i]));
      sta.emplace_back(i);
    }
    LCADoubling lca_g(g);
    lca_g.build(0);
    vector<pair<int, int>> dist(x, {n + 1, x});
    using P = pair<pair<int, int>, int>;
    priority_queue<P, vector<P>, greater<P>> que;
    for (int i = (0); i < (k); ++i) {
      dist[mp[v[i]]] = {0, i};
      que.emplace(dist[mp[v[i]]], mp[v[i]]);
    }
    while (!que.empty()) {
      auto [turn, type] = que.top().first;
      int ver = que.top().second;
      que.pop();
      if (dist[ver].first < turn) continue;
      if (dist[ver].second < type) continue;
      for (const Edge &e : g[ver]) {
        int tm = (lca_g.distance(mp[v[type]], e.dst) + s[type] - 1) / s[type];
        if (tm < dist[e.dst].first ||
            (tm == dist[e.dst].first && type < dist[e.dst].second)) {
          dist[e.dst] = {tm, type};
          que.emplace(dist[e.dst], e.dst);
        }
      }
    }
    for (int i = (0); i < (m); ++i)
      cout << dist[mp[u[i]]].second + 1 << " \n"[i + 1 == m];
  }
  return 0;
}