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 maxn = 200005, logn = 20;
int N, Q, in[maxn], out[maxn], st[maxn][logn], depth[maxn], cnt, spd[maxn],
    res[maxn], upd[maxn];
pair<int, int> dis[maxn];
vector<int> g[maxn], tmp[maxn];
void dfs(int u, int p) {
  in[u] = ++cnt;
  for (int v : g[u]) {
    if (v == p) continue;
    st[v][0] = u;
    depth[v] = depth[u] + 1;
    dfs(v, u);
  }
  out[u] = cnt;
}
int lca(int a, int b) {
  if (depth[a] > depth[b]) swap(a, b);
  int diff = depth[b] - depth[a];
  for (int i = 0; i < logn; i++)
    if (diff & (1 << i)) b = st[b][i];
  if (a == b) return a;
  for (int i = logn - 1; i >= 0; i--)
    if (st[a][i] != st[b][i]) a = st[a][i], b = st[b][i];
  return st[a][0];
}
inline void build(vector<int> &v, vector<int> &c) {
  vector<int> tot, t;
  for (int ck : v) tot.push_back(ck);
  for (int ck : c) tot.push_back(ck);
  sort(tot.begin(), tot.end());
  tot.resize(unique(tot.begin(), tot.end()) - tot.begin());
  sort(tot.begin(), tot.end(), [](int a, int b) { return in[a] < in[b]; });
  t = {tot[0]};
  for (int i = 1; i < tot.size(); i++) {
    t.push_back(lca(tot[i - 1], tot[i]));
    t.push_back(tot[i]);
  }
  sort(t.begin(), t.end());
  t.resize(unique(t.begin(), t.end()) - t.begin());
  sort(t.begin(), t.end(), [](int a, int b) { return in[a] < in[b]; });
  for (int v : t) tmp[v].clear();
  vector<int> s;
  for (int i = 0; i < t.size(); i++) {
    while (s.size() && out[s.back()] < in[t[i]]) s.pop_back();
    if (s.size()) {
      tmp[s.back()].push_back(t[i]);
      tmp[t[i]].push_back(s.back());
    }
    s.push_back(t[i]);
  }
}
inline int calc(pair<int, int> a) {
  return (a.first + spd[a.second] - 1) / spd[a.second];
}
inline bool better(pair<int, int> a, pair<int, int> b) {
  int f = calc(a), s = calc(b);
  if (f == s) return a.second < b.second;
  return f < s;
}
struct cmp {
  bool operator()(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
    if (a.first == b.first) return a.second > b.second;
    return !better(a.first, b.first);
  }
};
void solve(vector<int> &vir) {
  priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int>>,
                 cmp>
      q;
  ++cnt;
  for (int i = 1; i <= vir.size(); i++) {
    upd[vir[i - 1]] = cnt;
    res[vir[i - 1]] = i;
    dis[vir[i - 1]] = make_pair(0, i);
    q.push({{0, i}, vir[i - 1]});
  }
  while (q.size()) {
    pair<int, int> t = q.top().first;
    int node = q.top().second;
    q.pop();
    if (better(dis[node], t)) continue;
    for (int v : tmp[node]) {
      pair<int, int> nxt =
          make_pair(t.first + abs(depth[node] - depth[v]), t.second);
      if (upd[v] != cnt || better(nxt, dis[v])) {
        upd[v] = cnt;
        dis[v] = nxt;
        q.push(make_pair(dis[v], v));
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 1; i < N; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1, -1);
  for (int sz = 1; (1 << sz) <= N; sz++)
    for (int i = 1; i <= N; i++) st[i][sz] = st[st[i][sz - 1]][sz - 1];
  cin >> Q;
  while (Q--) {
    int vir, m;
    cin >> vir >> m;
    vector<int> v, cit;
    for (int i = 1; i <= vir; i++) {
      int a;
      cin >> a >> spd[i];
      v.push_back(a);
    }
    for (int i = 1; i <= m; i++) {
      int a;
      cin >> a;
      cit.push_back(a);
    }
    build(v, cit);
    solve(v);
    for (int ck : cit) cout << dis[ck].second << " ";
    cout << '\n';
  }
}