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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
using namespace std;
const int N = 200000 + 10;
const long long MOD = 1000000000 + 7;
const long long INF = 1000000010;
const long long LOG = 20;
int n, st[N], ft[N], Time, h[N], par[N][LOG], k, m, V[N], second[N], M[N];
vector<int> G[N];
vector<pair<int, int>> g[N];
pair<int, int> DIS[N];
bool cmp(int v, int u) { return st[v] < st[u]; }
void DFS(int v = 1, int p = 0) {
  st[v] = ++Time;
  par[v][0] = p;
  for (int i = 1; i < LOG; i++) par[v][i] = par[par[v][i - 1]][i - 1];
  for (auto u : G[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    DFS(u, v);
  }
  ft[v] = Time;
}
int LCA(int v, int u) {
  if (h[v] < h[u]) swap(v, u);
  int res = h[v] - h[u];
  for (int i = 0; i < LOG; i++)
    if (res & (1 << i)) v = par[v][i];
  if (v == u) return v;
  for (int i = LOG - 1; i >= 0; i--)
    if (par[v][i] != par[u][i]) v = par[v][i], u = par[u][i];
  return par[v][0];
}
int dis(int v, int u) {
  int lca = LCA(v, u);
  return h[v] + h[u] - 2 * h[lca];
}
int mod(int a) {
  a %= k;
  if (a < 0) a += k;
  return a;
}
int Ceil(int a, int b) {
  if (a % b == 0) return a / b;
  return a / b + 1;
}
int Main() {
  cin >> k >> m;
  for (int i = 0; i < k; i++) cin >> V[i] >> second[i];
  for (int i = 1; i <= m; i++) cin >> M[i];
  vector<int> nd;
  nd.push_back(1);
  for (int i = 0; i < k; i++) nd.push_back(V[i]);
  for (int i = 1; i <= m; i++) nd.push_back(M[i]);
  sort(nd.begin(), nd.end());
  nd.resize(unique(nd.begin(), nd.end()) - nd.begin());
  sort(nd.begin(), nd.end(), cmp);
  int SZ = nd.size();
  for (int i = 0; i < SZ - 1; i++) nd.push_back(LCA(nd[i], nd[i + 1]));
  sort(nd.begin(), nd.end());
  nd.resize(unique(nd.begin(), nd.end()) - nd.begin());
  sort(nd.begin(), nd.end(), cmp);
  for (auto u : nd) g[u].clear(), g[u].shrink_to_fit();
  stack<int> St;
  St.push(1);
  for (int i = 1; i < nd.size(); i++) {
    while (ft[St.top()] < st[nd[i]]) St.pop();
    int Dis = dis(St.top(), nd[i]);
    g[St.top()].push_back({nd[i], Dis}), g[nd[i]].push_back({St.top(), Dis});
    St.push(nd[i]);
  }
  set<pair<pair<int, int>, int>> pq;
  for (auto u : nd) DIS[u] = {INF, INF};
  for (int i = 0; i < k; i++) {
    DIS[V[i]] = {i - k, 0};
    pq.insert({DIS[V[i]], V[i]});
  }
  while (pq.size()) {
    auto fr = *pq.begin();
    pq.erase(fr);
    for (auto u : g[fr.second]) {
      pair<int, int> res;
      if (u.second <= fr.first.second)
        res = {fr.first.first, fr.first.second - u.second};
      else {
        int len = u.second;
        len -= fr.first.second;
        res.first =
            fr.first.first + (Ceil(len, second[mod(fr.first.first)]) * k);
        if (len % second[mod(fr.first.first)] == 0)
          res.second = 0;
        else {
          res.second =
              second[mod(fr.first.first)] - (len % second[mod(fr.first.first)]);
        }
      }
      if (DIS[u.first].first > res.first) {
        pq.erase({DIS[u.first], u.first});
        DIS[u.first] = res;
        pq.insert({DIS[u.first], u.first});
      } else if (DIS[u.first].first == res.first &&
                 DIS[u.first].second < res.second) {
        pq.erase({DIS[u.first], u.first});
        DIS[u.first] = res;
        pq.insert({DIS[u.first], u.first});
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << mod(DIS[M[i]].first) + 1 << ' ';
  }
  cout << '\n';
  return 0;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    G[v].push_back(u), G[u].push_back(v);
  }
  DFS();
  int q;
  cin >> q;
  while (q--) Main();
  return 0;
}