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 N = 2e5 + 7, K = 20, RT = 500;
vector<int> adj[N];
int st[N], en[N], dep[N], anc[N][K], order[N];
int tim = 0;
void dfs(int u, int p, int dep) {
  ::dep[u] = dep;
  st[u] = ++tim;
  order[tim] = u;
  anc[u][0] = p;
  for (int k = 1; k < K; k++) anc[u][k] = anc[anc[u][k - 1]][k - 1];
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs(v, u, dep + 1);
  }
  en[u] = tim;
}
int lca(int u, int v) {
  if (dep[u] > dep[v]) swap(u, v);
  for (int k = K - 1; k >= 0; k--)
    if (dep[u] + (1 << k) <= dep[v]) v = anc[v][k];
  if (u == v) return u;
  for (int k = K - 1; k >= 0; k--)
    if (anc[u][k] != anc[v][k]) u = anc[u][k], v = anc[v][k];
  return anc[u][0];
}
vector<pair<int, int>> ej[N];
pair<int, int> dis[N];
bool issub(int p, int u) { return st[p] <= st[u] && en[u] <= en[p]; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0, 0);
  int q;
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> special;
    vector<pair<int, int>> sources(k);
    for (int i = 0; i < k; i++) {
      cin >> sources[i].first >> sources[i].second;
      special.push_back(sources[i].first);
    }
    vector<int> qq(m);
    for (int i = 0; i < m; i++) {
      cin >> qq[i];
      special.push_back(qq[i]);
    }
    sort(special.begin(), special.end(),
         [](int u, int v) { return st[u] < st[v]; });
    special.erase(unique(special.begin(), special.end()), special.end());
    for (int i = special.size() - 1; i > 0; i--) {
      int a = special[i - 1];
      int b = special[i];
      special.push_back(lca(a, b));
    }
    sort(special.begin(), special.end(),
         [](int u, int v) { return st[u] < st[v]; });
    special.erase(unique(special.begin(), special.end()), special.end());
    for (int x : special) {
      ej[x].clear();
      dis[x] = {N, N};
    }
    stack<int> stk;
    stk.push(special[0]);
    for (int i = 1; i < special.size(); i++) {
      while (!issub(stk.top(), special[i])) stk.pop();
      int u = stk.top(), v = special[i];
      int d = dep[v] - dep[u];
      ej[u].push_back({v, d});
      ej[v].push_back({u, d});
      stk.push(special[i]);
    }
    priority_queue<tuple<int, int, int, int>> pq;
    for (int i = 0; i < sources.size(); i++) {
      int s = sources[i].first;
      dis[s] = {0, i};
      pq.push({-0, -i, s, 0});
    }
    while (pq.size()) {
      int k = -get<0>(pq.top());
      int i = -get<1>(pq.top());
      int u = get<2>(pq.top());
      int dd = get<3>(pq.top());
      pq.pop();
      if (make_pair(k, i) > dis[u]) continue;
      int s = sources[i].first;
      int sk = sources[i].second;
      for (auto pr : ej[u]) {
        int v = pr.first;
        int d = dd + pr.second;
        int kk = (d + sk - 1) / sk;
        if (make_pair(kk, i) < dis[v]) {
          dis[v] = make_pair(kk, i);
          pq.push({-kk, -i, v, d});
        }
      }
    }
    for (int x : qq) {
      cout << dis[x].second + 1 << " ";
    }
    cout << "\n";
  }
}