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;
int n, l = 18;
vector<int> g[N];
vector<pair<int, int>> gc[N];
int timer = 0;
int tin[N], tout[N];
int up[N][19];
int h[N], tr[N], ans[N];
const int INF = 1 << 30;
using ll = long long;
using Tp = tuple<int, int, int>;
using pp = pair<Tp, int>;
Tp dd[N];
void dfs(int v, int p) {
  if (p == v)
    h[v] = 0;
  else
    h[v] = h[p] + 1;
  tin[v] = ++timer;
  up[v][0] = p;
  for (int i = 1; i <= l; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  for (auto &u : g[v])
    if (u != p) dfs(u, v);
  tout[v] = ++timer;
}
bool is_ancestor(int u, int v) {
  return tin[u] <= tin[v] && tout[u] >= tout[v];
}
int lca(int u, int v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (int i = l; i >= 0; --i) {
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  }
  return up[u][0];
}
int dist(int u, int v) { return h[u] + h[v] - 2 * h[lca(u, v)]; }
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  for (int i = 1; i <= n; ++i) dd[i] = Tp(INF, INF, INF);
  dfs(1, 1);
  int q;
  cin >> q;
  auto cmp = [&](int a, int b) { return tin[a] < tin[b]; };
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> vir(k), spd(k);
    for (int i = 0; i < k; ++i) cin >> vir[i] >> spd[i];
    vector<int> imp(m);
    for (int i = 0; i < m; ++i) cin >> imp[i];
    vector<int> mrg = vir;
    mrg.insert(mrg.end(), imp.begin(), imp.end());
    sort(mrg.begin(), mrg.end(), cmp);
    mrg.erase(unique(mrg.begin(), mrg.end()), mrg.end());
    int x = mrg.size();
    for (int i = 0; i < x - 1; ++i) {
      mrg.emplace_back(lca(mrg[i], mrg[i + 1]));
    }
    sort(mrg.begin(), mrg.end(), cmp);
    mrg.erase(unique(mrg.begin(), mrg.end()), mrg.end());
    x = mrg.size();
    vector<int> st{mrg.front()};
    for (int i = 1; i < x; ++i) {
      while (!st.empty() && !is_ancestor(st.back(), mrg[i])) st.pop_back();
      auto a = st.back();
      auto b = mrg[i];
      auto l = dist(a, b);
      gc[a].emplace_back(b, l);
      gc[b].emplace_back(a, l);
      st.emplace_back(mrg[i]);
    }
    priority_queue<pp, vector<pp>, greater<pp>> qu;
    for (int i = 0; i < k; i++) {
      auto t = Tp(0, i, 0);
      dd[vir[i]] = t;
      qu.emplace(move(t), vir[i]);
    }
    while (!qu.empty()) {
      auto [nxt, city] = qu.top();
      qu.pop();
      if (dd[city] != nxt) continue;
      auto &[rnd, idx, len] = nxt;
      for (auto &v : gc[city]) {
        auto nxt_dist = len + v.second;
        auto t = Tp((nxt_dist + spd[idx] - 1) / spd[idx], idx, nxt_dist);
        if (dd[v.first] > t) {
          dd[v.first] = t;
          qu.emplace(move(t), v.first);
        }
      }
    }
    for (auto &v : imp) {
      cout << get<1>(dd[v]) + 1 << ' ';
    }
    cout << '\n';
    for (auto &a : mrg) {
      dd[a] = Tp(INF, INF, INF);
      gc[a].clear();
    }
  }
}