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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
const int MAXN = 2e5 + 7;
const int INF = 0x3f3f3f3f;
int n, q, k, m, dfn[MAXN], depth[MAXN], sz[MAXN], par[MAXN][20], num,
    dist[MAXN];
pair<int, int> sta[MAXN];
vector<int> G[MAXN], ipt;
vector<pair<int, int>> nG[MAXN], virus;
set<int> S;
void dfs(int u, int _par) {
  sz[u] = 1;
  depth[u] = depth[par[u][0] = _par] + 1;
  dfn[u] = ++num;
  for (int i = 1; par[u][i - 1]; i++) par[u][i] = par[par[u][i - 1]][i - 1];
  for (int v : G[u]) {
    if (v == _par) continue;
    dfs(v, u);
    sz[u] += sz[v];
  }
}
inline int LCA(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int i = 0; depth[u] - depth[v]; i++)
    if ((depth[u] - depth[v]) & (1 << i)) u = par[u][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; i--)
    if (par[u][i] != par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  return par[u][0];
}
void build(int &ptr, const vector<int> &vec, const int _par) {
  while (ptr < (int)vec.size()) {
    int u = vec[ptr];
    if (dfn[u] >= dfn[_par] + sz[_par]) return;
    nG[u].emplace_back(make_pair(_par, depth[u] - depth[_par]));
    nG[_par].emplace_back(make_pair(u, depth[u] - depth[_par]));
    ptr++;
    build(ptr, vec, u);
  }
}
void giao() {
  priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int>>,
                 greater<pair<pair<int, int>, int>>>
      que;
  for (int i = 0; i < (int)virus.size(); i++) {
    sta[virus[i].first] = make_pair(0, i);
    que.push(make_pair(make_pair(0, i), virus[i].first));
  }
  while (!que.empty()) {
    auto p = que.top();
    que.pop();
    if (p.first != sta[p.second]) continue;
    int u = p.second;
    int vID = p.first.second;
    for (auto e : nG[u]) {
      int v = e.first;
      int w = e.second;
      int nt = (dist[u] + w - 1) / virus[vID].second + 1;
      if (nt < sta[v].first || (nt == sta[v].first && vID < sta[v].second)) {
        sta[v].first = nt;
        sta[v].second = vID;
        dist[v] = dist[u] + w;
        que.push(make_pair(sta[v], v));
      }
    }
  }
}
void solve() {
  cin >> k >> m;
  vector<int> vec;
  ipt.clear();
  virus.resize(k);
  vec.resize(k + m);
  for (int i = 0; i < k; i++) {
    cin >> virus[i].first >> virus[i].second;
    vec[i] = virus[i].first;
  }
  for (int i = k; i < k + m; i++) {
    cin >> vec[i];
    ipt.emplace_back(vec[i]);
  }
  sort(vec.begin(), vec.end(),
       [](const int &A, const int &B) { return dfn[A] < dfn[B]; });
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  S.clear();
  for (int x : vec) S.insert(x);
  for (int i = 1; i < k + m; i++) {
    int lca = LCA(vec[i], vec[i - 1]);
    if (!S.count(lca)) {
      S.insert(lca);
      vec.emplace_back(lca);
    }
  }
  if (!S.count(1)) vec.emplace_back(1);
  sort(vec.begin(), vec.end(),
       [](const int &A, const int &B) { return dfn[A] < dfn[B]; });
  for (int x : vec) {
    nG[x].clear();
    sta[x] = make_pair(INF, INF);
    dist[x] = 0;
  }
  int ptr = 1;
  build(ptr, vec, 1);
  giao();
  for (int x : ipt) cout << sta[x].second + 1 << ' ';
  cout << endl;
}
int main() {
  ____();
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  dfs(1, 0);
  cin >> q;
  while (q--) solve();
  return 0;
}