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_MAX = 2e5;
int n, e[2 * N_MAX], h[N_MAX], first[N_MAX], nxt_e = 1;
int lg[2 * N_MAX], st[2 * N_MAX][20], L;
vector<int> G[N_MAX];
int q, k, m, virus[N_MAX], speed[N_MAX], city[N_MAX], in_all[N_MAX];
vector<int> all;
vector<pair<int, int>> T[N_MAX];
stack<int> s;
void DFS(int u) {
  for (int v : G[u]) {
    if (first[v] != -1) continue;
    h[v] = h[u] + 1;
    first[v] = nxt_e;
    e[nxt_e++] = v;
    DFS(v);
    e[nxt_e++] = u;
  }
}
int lca(int u, int v) {
  u = first[u];
  v = first[v];
  if (u > v) swap(u, v);
  int l = lg[v - u + 1];
  int x = st[u][l], y = st[v - (1 << l) + 1][l];
  return (h[x] < h[y] ? x : y);
}
tuple<int, int, int> d[N_MAX];
priority_queue<pair<tuple<int, int, int>, int>,
               vector<pair<tuple<int, int, int>, int>>,
               greater<pair<tuple<int, int, int>, int>>>
    pq;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  for (int i = 0; i < n; i++) first[i] = -1;
  h[0] = e[0] = first[0] = 0;
  DFS(0);
  for (int i = 1; i <= 2 * n - 1; i++) {
    if ((1 << (L + 1)) <= i) L++;
    lg[i] = L;
  }
  for (int i = 0; i < 2 * n - 1; i++) st[i][0] = e[i];
  for (int l = 1; l <= L; l++) {
    for (int i = 0; i < 2 * n - 1; i++) {
      st[i][l] = st[i][l - 1];
      if (i + (1 << (l - 1)) < 2 * n - 1) {
        int v = st[i + (1 << (l - 1))][l - 1];
        if (h[v] < h[st[i][l]]) st[i][l] = v;
      }
    }
  }
  cin >> q;
  for (int t = 1; t <= q; t++) {
    cin >> k >> m;
    all.clear();
    for (int i = 0; i < k; i++) {
      cin >> virus[i] >> speed[i];
      all.push_back(--virus[i]);
      in_all[virus[i]] = t;
    }
    for (int i = 0; i < m; i++) {
      cin >> city[i];
      city[i]--;
      if (in_all[city[i]] != t) {
        all.push_back(city[i]);
        in_all[city[i]] = t;
      }
    }
    sort(all.begin(), all.end(), [](const int& u, const int& v) -> bool {
      return first[u] < first[v];
    });
    int all_size = all.size();
    for (int i = 0; i + 1 < all_size; i++) {
      int v = lca(all[i], all[i + 1]);
      if (in_all[v] != t) {
        all.push_back(v);
        in_all[v] = t;
      }
    }
    sort(all.begin(), all.end(), [](const int& u, const int& v) -> bool {
      return first[u] < first[v];
    });
    for (int v : all) T[v].clear();
    stack<int>().swap(s);
    s.push(all[0]);
    for (int i = 1; i < all.size(); i++) {
      while (lca(all[i], s.top()) != s.top()) s.pop();
      int d = h[all[i]] - h[s.top()];
      T[s.top()].push_back({all[i], d});
      T[all[i]].push_back({s.top(), d});
      s.push(all[i]);
    }
    priority_queue<pair<tuple<int, int, int>, int>,
                   vector<pair<tuple<int, int, int>, int>>,
                   greater<pair<tuple<int, int, int>, int>>>()
        .swap(pq);
    for (int u : all) get<0>(d[u]) = -1;
    for (int i = 0; i < k; i++) {
      d[virus[i]] = make_tuple(0, i, speed[i]);
      pq.push(make_pair(d[virus[i]], virus[i]));
    }
    while (!pq.empty()) {
      auto [tmp, u] = pq.top();
      pq.pop();
      if (d[u] != tmp) continue;
      for (auto [v, w] : T[u]) {
        tuple<int, int, int> r = d[u];
        int sp = speed[get<1>(r)];
        if (get<2>(r) + w <= sp)
          get<2>(r) += w;
        else {
          w -= sp - get<2>(r);
          get<0>(r) += (w + sp - 1) / sp;
          get<2>(r) = w % sp;
          if (get<2>(r) == 0) get<2>(r) = sp;
        }
        if (get<0>(d[v]) == -1 || r < d[v]) {
          d[v] = r;
          pq.push(make_pair(d[v], v));
        }
      }
    }
    for (int i = 0; i < m; i++) cout << get<1>(d[city[i]]) + 1 << " ";
    cout << "\n";
  }
  return 0;
}