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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = (int)2e5 + 10;
const int LOG = 19;
vector<int> T[N];
vector<pair<int, int> > G[N];
int tin[N];
int tout[N];
int par[LOG][N];
int dis[N];
int ti;
void dfs(int u, int pr) {
  par[0][u] = pr;
  for (int i = 1; i < LOG; i++) {
    par[i][u] = par[i - 1][par[i - 1][u]];
  }
  tin[u] = ++ti;
  for (auto x : T[u]) {
    if (x == pr) continue;
    dis[x] = dis[u] + 1;
    dfs(x, u);
  }
  tout[u] = ti;
}
bool is_anc(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (is_anc(a, b)) return a;
  if (is_anc(b, a)) return b;
  for (int i = LOG - 1; i >= 0; i--) {
    if (!is_anc(par[i][a], b)) a = par[i][a];
  }
  return par[0][a];
}
vector<pair<int, int> > vv;
vector<int> imp;
vector<pair<int, int> > nod;
struct S {
  int node;
  int id;
  int dist;
  bool operator<(const S &F) const {
    if ((dist + vv[id].second - 1) / vv[id].second ==
        (F.dist + vv[F.id].second - 1) / vv[F.id].second) {
      if (id == F.id) return node < F.node;
      return id < F.id;
    }
    return (dist + vv[id].second - 1) / vv[id].second <
           (F.dist + vv[F.id].second - 1) / vv[F.id].second;
  }
};
S F[N];
void solve() {
  vv.clear();
  imp.clear();
  nod.clear();
  int k, m;
  cin >> k >> m;
  int a, b;
  for (int i = 0; i < k; i++) {
    cin >> a >> b;
    vv.push_back(make_pair(a, b));
    nod.push_back(make_pair(tin[a], a));
  }
  for (int i = 0; i < m; i++) {
    cin >> a;
    imp.push_back(a);
    nod.push_back(make_pair(tin[a], a));
  }
  int len;
  sort(nod.begin(), nod.end());
  nod.resize(unique(nod.begin(), nod.end()) - nod.begin());
  int l = nod.size();
  int v;
  for (int i = 0; i + 1 < l; i++) {
    v = lca(nod[i].second, nod[i + 1].second);
    nod.push_back(make_pair(tin[v], v));
  }
  sort(nod.begin(), nod.end());
  nod.resize(unique(nod.begin(), nod.end()) - nod.begin());
  vector<int> cur;
  for (auto p : nod) {
    while (!cur.empty() && !is_anc(cur.back(), p.second)) cur.pop_back();
    if (!cur.empty()) {
      len = dis[p.second] - dis[cur.back()];
      G[p.second].push_back(make_pair(cur.back(), len));
      G[cur.back()].push_back(make_pair(p.second, len));
    }
    cur.push_back(p.second);
  }
  for (auto x : nod) {
    F[x.second] = {x.second, -1, -1};
  }
  set<S> pq;
  for (int i = 0; i < vv.size(); i++) {
    F[vv[i].first] = {vv[i].first, i, 0};
    pq.insert({vv[i].first, i, 0});
  }
  S cr, nx;
  while (!pq.empty()) {
    cr = *pq.begin();
    pq.erase(pq.begin());
    for (auto x : G[cr.node]) {
      nx = {x.first, cr.id, cr.dist + x.second};
      if (F[x.first].id == -1) {
        F[x.first] = nx;
        pq.insert(nx);
      } else if (nx < F[x.first]) {
        pq.erase(F[x.first]);
        F[x.first] = nx;
        pq.insert(nx);
      }
    }
  }
  for (auto v : imp) cout << F[v].id + 1 << " ";
  cout << "\n";
  for (auto x : nod) {
    G[x.second].clear();
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    T[u].push_back(v);
    T[v].push_back(u);
  }
  dfs(1, 1);
  int q;
  cin >> q;
  for (int t = 0; t < q; t++) solve();
  return 0;
}