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 inf = 1e9 + 7;
const int N = 200050;
const int L = 20;
vector<int> E[N];
vector<pair<int, int>> G[N];
int lid[N], rid[N], tid, dep[N], par[N][L];
void DFS(int u, int p) {
  lid[u] = ++tid;
  dep[u] = dep[p] + 1;
  par[u][0] = p;
  for (int i = 1; i < L; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  for (int v : E[u])
    if (v != p) DFS(v, u);
  rid[u] = tid;
}
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = L - 1; ~i; i--)
    if (dep[par[u][i]] >= dep[v]) u = par[u][i];
  for (int i = L - 1; ~i; i--)
    if (par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
  return u == v ? v : par[v][0];
}
void AddEdge(int u, int v, int w) {
  G[u].push_back({v, w});
  G[v].push_back({u, w});
}
struct triple {
  int f, s, t;
  triple() {}
  triple(int a, int b, int c) : f(a), s(b), t(c) {}
} dist[N];
bool operator<(triple a, triple b) {
  return make_tuple(b.f, b.s, b.t) < make_tuple(a.f, a.s, a.t);
}
bool operator!=(triple a, triple b) {
  return make_tuple(b.f, b.s, b.t) != make_tuple(a.f, a.s, a.t);
}
int main() {
  int n;
  scanf("%i", &n);
  for (int i = 1, u, v; i < n; i++)
    scanf("%i %i", &u, &v), E[u].push_back(v), E[v].push_back(u);
  DFS(1, 0);
  int q;
  scanf("%i", &q);
  while (q--) {
    int k, m;
    scanf("%i %i", &k, &m);
    vector<pair<int, int>> vir(k);
    vector<int> imp(m), nds;
    for (auto &i : vir)
      scanf("%i %i", &i.first, &i.second), nds.push_back(i.first);
    for (auto &i : imp) scanf("%i", &i), nds.push_back(i);
    sort(nds.begin(), nds.end(), [&](int i, int j) { return lid[i] < lid[j]; });
    for (int i = 1; i < m + k; i++) nds.push_back(LCA(nds[i - 1], nds[i]));
    sort(nds.begin(), nds.end(), [&](int i, int j) { return lid[i] < lid[j]; });
    nds.erase(unique(nds.begin(), nds.end()), nds.end());
    stack<int> stk;
    for (int u : nds) {
      if (stk.empty())
        stk.push(u);
      else {
        while (rid[stk.top()] < lid[u]) stk.pop();
        AddEdge(stk.top(), u, dep[u] - dep[stk.top()]);
        stk.push(u);
      }
    }
    for (int u : nds) dist[u] = triple(inf, 0, 0);
    priority_queue<pair<triple, int>> pq;
    for (int i = 0; i < k; i++) {
      dist[vir[i].first] = {0, i, 0};
      pq.push({dist[vir[i].first], vir[i].first});
    }
    while (pq.size()) {
      int u = pq.top().second;
      triple d = pq.top().first;
      pq.pop();
      if (d != dist[u]) continue;
      for (auto e : G[u]) {
        triple d2 = d;
        d2.t += e.second;
        d2.f = (d2.t + vir[d2.s].second - 1) / vir[d2.s].second;
        if (dist[e.first] < d2) {
          dist[e.first] = d2;
          pq.push({d2, e.first});
        }
      }
    }
    for (int u : imp) printf("%i ", dist[u].s + 1);
    printf("\n");
    for (int u : nds) G[u].clear();
  }
  return 0;
}