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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const int N = 200005, LOG = 18;
int n, k;
int a, b;
vector<int> V[N];
bool vis[N];
int inTime[N], outTime[N], timer;
int par[N][LOG + 5];
int parent[N];
int depth[N];
vector<pair<int, int>> G[N];
int dist[N], source[N];
void DFS(int w) {
  inTime[w] = ++timer;
  vis[w] = true;
  par[w][0] = parent[w];
  for (int i = 1; i < LOG; i++) {
    par[w][i] = par[par[w][i - 1]][i - 1];
  }
  for (int u : V[w]) {
    if (!vis[u]) {
      parent[u] = w;
      depth[u] = depth[w] + 1;
      DFS(u);
    }
  }
  outTime[w] = ++timer;
}
bool isAncestor(int anc, int w) {
  return inTime[anc] <= inTime[w] && outTime[anc] >= outTime[w];
}
int LCA(int a, int b) {
  for (int i = LOG - 1; i >= 0; i--) {
    if (depth[a] - (1 << i) >= depth[b]) {
      a = par[a][i];
    }
  }
  for (int i = LOG - 1; i >= 0; i--) {
    if (depth[b] - (1 << i) >= depth[a]) {
      b = par[b][i];
    }
  }
  if (a == b) {
    return a;
  }
  for (int i = LOG - 1; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      a = par[a][i];
      b = par[b][i];
    }
  }
  return par[a][0];
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d %d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  DFS(1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int virNo, nodesNo;
    scanf("%d %d", &virNo, &nodesNo);
    vector<pair<int, int>> viruses;
    vector<int> queryNodes;
    vector<int> nodes;
    for (int i = 0; i < virNo; i++) {
      int init, speed;
      scanf("%d %d", &init, &speed);
      viruses.push_back({init, speed});
      nodes.push_back(init);
    }
    for (int i = 0; i < nodesNo; i++) {
      int v;
      scanf("%d", &v);
      queryNodes.push_back(v);
      nodes.push_back(v);
    }
    sort(nodes.begin(), nodes.end(),
         [&](int a, int b) { return inTime[a] < inTime[b]; });
    for (int i = nodes.size() - 1; i > 0; i--) {
      nodes.push_back(LCA(nodes[i], nodes[i - 1]));
    }
    sort(nodes.begin(), nodes.end(),
         [&](int a, int b) { return inTime[a] < inTime[b]; });
    nodes.resize(unique(nodes.begin(), nodes.end()) - nodes.begin());
    vector<int> s;
    for (int i = 0; i < nodes.size(); i++) {
      while (!s.empty() && !isAncestor(s.back(), nodes[i])) {
        s.pop_back();
      }
      if (!s.empty()) {
        ("krawedz", s.back(), nodes[i], depth[nodes[i]] - depth[s.back()]);
        G[s.back()].push_back({nodes[i], depth[nodes[i]] - depth[s.back()]});
        G[nodes[i]].push_back({s.back(), depth[nodes[i]] - depth[s.back()]});
      }
      s.push_back(nodes[i]);
    }
    for (int i = 0; i < nodes.size(); i++) {
      dist[nodes[i]] = 1e9;
      vis[nodes[i]] = false;
    }
    priority_queue<pair<pair<int, int>, int>> Q;
    for (int i = 0; i < viruses.size(); i++) {
      dist[viruses[i].first] = i;
      source[viruses[i].first] = i;
      Q.push({{-i, viruses[i].first}, 0});
    }
    while (!Q.empty()) {
      int w = Q.top().first.second;
      int curDist = Q.top().second;
      Q.pop();
      if (vis[w]) continue;
      vis[w] = true;
      for (auto p : G[w]) {
        int u = p.first;
        int kr = p.second;
        int realDist = curDist + kr;
        int myD = (curDist + viruses[source[w]].second - 1) /
                  viruses[source[w]].second;
        int sonD = (realDist + viruses[source[w]].second - 1) /
                   viruses[source[w]].second;
        int newDist = dist[w] + virNo * (sonD - myD);
        if (newDist < dist[u]) {
          dist[u] = newDist;
          source[u] = source[w];
          Q.push({{-dist[u], u}, realDist});
        }
      }
    }
    for (int u : queryNodes) {
      printf("%d ", source[u] + 1);
    }
    printf("\n");
    for (int w : nodes) {
      G[w].clear();
    }
  }
  return 0;
}