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;
template <typename T>
void cetak(T t) {
  cout << t << ')' << endl;
}
template <typename T, typename... V>
void cetak(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ", ";
  cetak(v...);
}
typedef int (*func)(int, int);
const int MOD = 1e9 + 7;
int add(int x, int y) {
  int ret = x + y;
  if (ret >= MOD) ret -= MOD;
  return ret;
}
int sub(int x, int y) {
  int ret = x - y;
  if (ret < 0) ret += MOD;
  return ret;
}
int mul(int x, int y) { return (long long)x * y % MOD; }
const int mx = 1e5 + 10;
struct LCA {
  int n, idx;
  const int LOG = 20;
  vector<int> in, out, level;
  vector<vector<int>> P, G;
  LCA() {
    cin >> n;
    P.resize(n, vector<int>(LOG, 0));
    G.resize(n);
    in.resize(n);
    out.resize(n);
    level.resize(n, 0);
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      u -= 1;
      v -= 1;
      G[v].push_back(u);
      G[u].push_back(v);
    }
    idx = -1;
    dfs(0, 0);
  }
  LCA(vector<vector<int>> g) {
    n = (int)(g.size());
    G = g;
    P.resize(n, vector<int>(LOG, 0));
    in.resize(n);
    out.resize(n);
    level.resize(n, 0);
    idx = -1;
    dfs(0, 0);
  }
  void dfs(int now, int par) {
    P[now][0] = par;
    for (int i = 1; i < LOG; i++) P[now][i] = P[P[now][i - 1]][i - 1];
    in[now] = ++idx;
    for (int i : G[now]) {
      if (i == par) continue;
      level[i] = level[now] + 1;
      dfs(i, now);
    }
    out[now] = idx;
  }
  int is_ancestor(int p, int q) { return in[p] <= in[q] && out[q] <= out[p]; }
  int query(int p, int q) {
    if (is_ancestor(p, q)) return p;
    if (is_ancestor(q, p)) return q;
    for (int i = LOG - 1; i >= 0; i--)
      if (!is_ancestor(P[p][i], q)) p = P[p][i];
    return P[p][0];
  }
  int dist(int p, int q) {
    return level[p] + level[q] - 2 * level[query(p, q)];
  }
};
vector<pair<int, int>> compress_tree(LCA &lca, vector<int> important) {
  sort(important.begin(), important.end(),
       [&](int i, int j) { return lca.in[i] < lca.in[j]; });
  important.erase(unique(important.begin(), important.end()), important.end());
  int siz = important.size();
  for (int i = 1; i < siz; i++) {
    important.push_back(lca.query(important[i], important[i - 1]));
  }
  sort(important.begin(), important.end(),
       [&](int i, int j) { return lca.in[i] < lca.in[j]; });
  important.erase(unique(important.begin(), important.end()), important.end());
  vector<pair<int, int>> ret;
  for (int i = 0; i < (int)(important.size()); i++) {
    if (i == 0)
      ret.push_back({important[i], important[i]});
    else
      ret.push_back({lca.query(important[i], important[i - 1]), important[i]});
  }
  return ret;
}
void solve() {
  int n, q;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u -= 1;
    v -= 1;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  LCA lca(g);
  vector<int> di(n), virus(n);
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> v(k), second(k), u(m);
    for (int i = 0; i < k; i++) {
      cin >> v[i] >> second[i];
      v[i] -= 1;
    }
    for (int &i : u) cin >> i, i -= 1;
    vector<int> gabung;
    for (int i : v) gabung.push_back(i);
    for (int i : u) gabung.push_back(i);
    auto ve = compress_tree(lca, gabung);
    int N = (int)(ve.size());
    vector<vector<int>> G(N);
    for (int i = 0; i < N; i++) {
      di[ve[i].second] = i;
      virus[i] = -1;
    }
    for (int i = 1; i < N; i++) {
      int x = di[ve[i].second];
      int p = di[ve[i].first];
      G[x].push_back(p);
      G[p].push_back(x);
    }
    priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int>>,
                   greater<pair<pair<int, int>, int>>>
        pq;
    for (int i = 0; i < k; i++) {
      pq.push({{0, i}, di[v[i]]});
    }
    while (!pq.empty()) {
      int i = pq.top().second;
      int id = pq.top().first.second;
      pq.pop();
      if (virus[i] == -1) {
        virus[i] = id;
        for (int j : G[i]) {
          int dist = lca.dist(v[id], ve[j].second);
          int kapan = (dist - 1) / second[id] + 1;
          pq.push({{kapan, id}, j});
        }
      }
    }
    for (int i : u) cout << virus[di[i]] + 1 << ' ';
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}