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 + 5;
int n, q, l[N], r[N], level[N], p[N][20], res[N];
int import[N], numCity, numVirus;
vector<int> adj[N];
vector<pair<int, int> > newAdj[N];
struct T {
  int city, speed;
};
T virus[N];
pair<pair<int, int>, int> top;
struct Less {
  bool operator()(const pair<pair<int, int>, int>& x,
                  const pair<pair<int, int>, int>& y) {
    int a = x.first.first / virus[x.first.second].speed,
        b = y.first.first / virus[y.first.second].speed;
    if (a != b) return a > b;
    return x.first.second > y.first.second;
  }
};
priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int> >,
               Less>
    pq;
int cnt = 0;
void dfs(int x) {
  cnt++;
  l[x] = cnt;
  for (int i = 1; i <= 18; i++) p[x][i] = p[p[x][i - 1]][i - 1];
  for (auto& u : adj[x]) {
    if (u == p[x][0]) continue;
    level[u] = level[x] + 1;
    p[u][0] = x;
    dfs(u);
  }
  r[x] = cnt;
}
int lca(int x, int y) {
  if (x == 1 || y == 1) return 1;
  if (level[x] > level[y]) swap(x, y);
  int diff = level[y] - level[x];
  for (int k = 18; k >= 0; k--)
    if (diff >> k & 1) y = p[y][k];
  if (x == y) return x;
  for (int k = 18; k >= 0; k--) {
    if (p[x][k] != p[y][k]) {
      x = p[x][k];
      y = p[y][k];
    }
  }
  return p[x][0];
}
int dist(int x, int y) { return level[x] + level[y] - 2 * level[lca(x, y)]; }
vector<pair<int, int> > all;
void compress_tree() {
  all.clear();
  for (int i = 1; i <= numVirus; i++)
    all.push_back(make_pair(l[virus[i].city], virus[i].city));
  for (int i = 1; i <= numCity; i++)
    all.push_back(make_pair(l[import[i]], import[i]));
  sort(all.begin(), all.end());
  all.erase(unique(all.begin(), all.end()), all.end());
  int sz = all.size();
  for (int i = 1; i < sz; i++) {
    int p = lca(all[i].second, all[i - 1].second);
    all.push_back(make_pair(l[p], p));
  }
  sort(all.begin(), all.end());
  all.erase(unique(all.begin(), all.end()), all.end());
  vector<int> st;
  for (auto& u : all) {
    newAdj[u.second].clear();
    while (st.size() && r[st.back()] < u.first) st.pop_back();
    if (st.size()) {
      int dist = level[u.second] - level[st.back()];
      newAdj[st.back()].push_back(make_pair(u.second, dist));
      newAdj[u.second].push_back(make_pair(st.back(), dist));
    }
    st.push_back(u.second);
  }
}
void solve() {
  for (auto& x : all) res[x.second] = -1;
  for (int i = 1; i <= numVirus; i++)
    pq.push({{virus[i].speed - 1, i}, virus[i].city});
  while (pq.size()) {
    top = pq.top();
    pq.pop();
    if (res[top.second] != -1) continue;
    res[top.second] = top.first.second;
    for (auto u : newAdj[top.second]) {
      if (res[u.first] != -1) continue;
      pq.push({{top.first.first + u.second, top.first.second}, u.first});
    }
  }
  for (int i = 1; i <= numCity; i++) cout << res[import[i]] << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  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);
  cin >> q;
  while (q--) {
    cin >> numVirus >> numCity;
    for (int i = 1; i <= numVirus; i++) cin >> virus[i].city >> virus[i].speed;
    for (int i = 1; i <= numCity; i++) cin >> import[i];
    compress_tree();
    solve();
  }
}