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 MX = 200005, LG = 20;
int n, q, f[MX], p[LG][MX], lvl[MX], cn = 1, from[MX], inv[MX];
int k, m, vid[MX], vs[MX], ic[MX], ini[MX], fin[MX];
priority_queue<pair<long long, int>, vector<pair<long long, int>>,
               greater<pair<long long, int>>>
    pq;
long long dis[MX], sz;
vector<int> gra[MX];
vector<pair<long long, int>> adj[MX];
void dfs(int u) {
  for (int v : gra[u])
    if (!f[v]) {
      f[v] = ++cn;
      p[0][f[v]] = f[u];
      lvl[f[v]] = lvl[f[u]] + 1;
      dfs(v);
    }
}
int LCA(int a, int b) {
  if (lvl[a] < lvl[b]) swap(a, b);
  for (int j = 0; j < LG; j++)
    if (lvl[a] - lvl[b] & (1 << j)) a = p[j][a];
  if (a == b) return a;
  for (int j = LG - 1; j + 1; j--)
    if (p[j][a] != p[j][b]) {
      a = p[j][a];
      b = p[j][b];
    }
  return p[0][a];
}
int kth_anc(int u, int k) {
  for (int j = 0; j < LG; j++)
    if (k & (1 << j)) u = p[j][u];
  return u;
}
void add(int u, int v, int w) {
  adj[u].emplace_back(v, w);
  adj[v].emplace_back(u, w);
}
void build(vector<int> &st) {
  stack<int> anc;
  for (int u : st) {
    adj[u].clear();
    from[u] = -1;
    while (anc.size()) {
      int v = anc.top();
      if (lvl[v] < lvl[u] && kth_anc(u, lvl[u] - lvl[v]) == v) break;
      anc.pop();
    }
    if (anc.size()) add(u, anc.top(), lvl[u] - lvl[anc.top()]);
    anc.push(u);
  }
}
void bfs() {
  sz = 0;
  for (int i = 0; i < k; i++) {
    int u = f[vid[i]];
    if (i) ini[i] = ini[i - 1] + vs[i - 1];
    fin[i] = ini[i] + vs[i] - 1;
    sz += vs[i];
    from[u] = i;
    dis[u] = ini[i] - 1;
    pq.emplace(dis[u], u);
  }
  while (pq.size()) {
    long long w = pq.top().first;
    int u = pq.top().second;
    int i = from[u];
    pq.pop();
    if (w > dis[u]) continue;
    for (pair<long long, int> &v : adj[u]) {
      long long d = w;
      long long r = v.second;
      if (r > fin[i] - d % sz) {
        r -= fin[i] - d % sz;
        d = sz * (max(d / sz, 0ll) + 1);
        long long q = r / (fin[i] - ini[i] + 1);
        r %= fin[i] - ini[i] + 1;
        d += q * sz;
        if (r)
          d += ini[i] + r - 1;
        else
          d += fin[i] - sz;
      } else {
        d += r;
      }
      if (from[v.first] == -1 ||
          (f[vid[from[v.first]]] != v.first && dis[v.first] > d)) {
        dis[v.first] = d;
        from[v.first] = i;
        pq.emplace(d, v.first);
      }
    }
  }
}
void solve() {
  vector<int> in, st;
  cin >> k >> m;
  for (int i = 0; i < k; i++) {
    cin >> vid[i] >> vs[i];
    in.push_back(f[vid[i]]);
    st.push_back(f[vid[i]]);
  }
  for (int i = 0; i < m; i++) {
    cin >> ic[i];
    in.push_back(f[ic[i]]);
    st.push_back(f[ic[i]]);
  }
  sort(in.begin(), in.end());
  in.erase(unique(in.begin(), in.end()), in.end());
  for (int i = 0; i < int(in.size()) - 1; i++)
    st.push_back(LCA(in[i], in[i + 1]));
  sort(st.begin(), st.end());
  st.erase(unique(st.begin(), st.end()), st.end());
  build(st);
  bfs();
  for (int i = 0; i < m; i++) cout << from[f[ic[i]]] + 1 << " ";
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    gra[a].push_back(b);
    gra[b].push_back(a);
  }
  f[1] = 1;
  p[0][1] = 1;
  dfs(1);
  for (int i = 1; i <= n; i++) inv[f[i]] = i;
  for (int j = 1; j < LG; j++)
    for (int i = 1; i <= n; i++) p[j][i] = p[j - 1][p[j - 1][i]];
  cin >> q;
  while (q--) solve();
  return 0;
}