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;
using ll = long long;
using pp = pair<pair<int, int>, int>;
const int maxn = 2e5 + 5, maxk = 18;
int n, q;
vector<int> adj[maxn];
int depth[maxn];
int par[maxk][maxn];
int tin[maxn], tout[maxn], t = 1;
void dfs(int i, int p) {
  tin[i] = t++;
  depth[i] = depth[p] + 1;
  par[0][i] = p;
  for (int k = 1; k < maxk; k++) {
    par[k][i] = par[k - 1][par[k - 1][i]];
  }
  for (int j : adj[i]) {
    if (j == p) continue;
    dfs(j, i);
  }
  tout[i] = t - 1;
}
int lca(int a, int b) {
  if (depth[a] > depth[b]) swap(a, b);
  for (int k = maxk - 1; k >= 0; k--) {
    int bb = par[k][b];
    if (depth[bb] >= depth[a]) b = bb;
  }
  if (a == b) return a;
  for (int k = maxk - 1; k >= 0; k--) {
    int aa = par[k][a];
    int bb = par[k][b];
    if (aa != bb) {
      a = aa;
      b = bb;
    }
  }
  return par[0][a];
}
int dis(int a, int b) { return depth[a] + depth[b] - 2 * depth[lca(a, b)]; }
vector<pair<int, int>> graph[maxn];
pair<int, int> dist[maxn];
vector<int> solve(vector<pair<int, int>> virus, vector<int> special) {
  vector<int> sp;
  for (auto p : virus) sp.push_back(p.first);
  for (auto i : special) sp.push_back(i);
  sort(sp.begin(), sp.end(), [](int a, int b) { return tin[a] < tin[b]; });
  sp.erase(unique(sp.begin(), sp.end()), sp.end());
  int nn = sp.size();
  for (int i = 1; i < nn; i++) {
    sp.push_back(lca(sp[i - 1], sp[i]));
  }
  sort(sp.begin(), sp.end(), [](int a, int b) { return tin[a] < tin[b]; });
  sp.erase(unique(sp.begin(), sp.end()), sp.end());
  for (auto i : sp) {
    graph[i].clear();
  }
  stack<int> st;
  for (auto i : sp) {
    while (!st.empty() && tin[i] > tout[st.top()]) {
      st.pop();
    }
    if (!st.empty()) {
      int len = depth[i] - depth[st.top()];
      graph[st.top()].push_back({i, len});
      graph[i].push_back({st.top(), len});
    }
    st.push(i);
  }
  priority_queue<pp, vector<pp>, greater<pp>> pq;
  for (auto i : sp) dist[i] = {maxn, -1};
  for (int i = 0; i < virus.size(); i++) {
    pq.push(make_pair(make_pair(0, i), virus[i].first));
    dist[virus[i].first] = {0, i};
  }
  while (!pq.empty()) {
    int u = pq.top().second;
    int s = virus[dist[u].second].second;
    int source = virus[dist[u].second].first;
    pq.pop();
    for (auto e : graph[u]) {
      int v = e.first;
      int nt = (dis(source, v) + s - 1) / s;
      if (dist[v] > make_pair(nt, dist[u].second)) {
        dist[v] = make_pair(nt, dist[u].second);
        pq.push(make_pair(dist[v], v));
      }
    }
  }
  vector<int> ret;
  for (auto i : special) {
    ret.push_back(dist[i].second);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<pair<int, int>> virus;
    vector<int> special;
    while (k--) {
      int v, s;
      cin >> v >> s;
      virus.push_back({v, s});
    }
    while (m--) {
      int v;
      cin >> v;
      special.push_back(v);
    }
    auto ans = solve(virus, special);
    for (auto i : ans) {
      cout << i + 1 << ' ';
    }
    cout << '\n';
  }
}