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 K = 18, MAXN = 1 << K;
vector<int> gr[MAXN];
vector<pair<int, int>> g[MAXN];
int F[K][MAXN], D[MAXN], idd, tim[MAXN], sz[MAXN], n;
pair<int, int> dist[MAXN];
void lca_dfs(int x) {
  for (int i = 0, ThxDem = gr[x].size(); i < ThxDem; ++i) {
    int y = gr[x][i];
    if (y == F[0][x]) continue;
    F[0][y] = x;
    D[y] = D[x] + 1;
    lca_dfs(y);
  }
}
void lca_init() {
  D[0] = 0;
  F[0][0] = -1;
  lca_dfs(0);
  for (int k = 1, ThxDem = K; k < ThxDem; ++k)
    for (int x = 0, ThxDem = n; x < ThxDem; ++x)
      if (F[k - 1][x] < 0)
        F[k][x] = -1;
      else
        F[k][x] = F[k - 1][F[k - 1][x]];
}
int lca(int x, int y) {
  if (D[x] < D[y]) swap(x, y);
  for (int k = K - 1; k >= 0; --k)
    if (D[x] - (1 << k) >= D[y]) x = F[k][x];
  if (x == y) return x;
  for (int k = K - 1; k >= 0; --k)
    if (F[k][x] != F[k][y]) x = F[k][x], y = F[k][y];
  return F[0][x];
}
bool cmp2(int x, int y) { return tim[x] < tim[y]; }
bool cmp(pair<int, int> x, pair<int, int> y) { return cmp2(x.first, y.first); }
void go(vector<pair<int, int>> &vir, vector<int> &spe) {
  vector<int> all = spe;
  for (auto x : vir) all.push_back(x.first);
  sort(all.begin(), all.end(), cmp2);
  for (int i = 1, ThxDem = int(all.size()); i < ThxDem; ++i)
    all.push_back(lca(all[i - 1], all[i]));
  sort(all.begin(), all.end(), cmp2);
  all.erase(unique(all.begin(), all.end()), all.end());
  for (auto x : all) g[x].clear(), dist[x] = {1e9, 1e9};
  stack<int> st;
  int root = -1;
  for (auto x : all) {
    while (int(st.size()) && tim[st.top()] + sz[st.top()] <= tim[x]) st.pop();
    if (int(st.size())) {
      int v = st.top(), ds = D[v] + D[x] - 2 * D[lca(x, v)];
      g[x].push_back({v, ds});
      g[v].push_back({x, ds});
    } else
      root = x;
    st.push(x);
  }
  priority_queue<pair<pair<int, int>, pair<int, int>>,
                 vector<pair<pair<int, int>, pair<int, int>>>,
                 greater<pair<pair<int, int>, pair<int, int>>>>
      q;
  for (auto x : vir)
    dist[x.first] = {0, int(q.size())},
    q.push({{0, dist[x.first].second}, {0, x.first}});
  while (int(q.size())) {
    int tr = q.top().first.first, id = q.top().first.second;
    int rem = q.top().second.first, wh = q.top().second.second,
        vel = vir[id].second;
    q.pop();
    if (dist[wh] != (pair<int, int>){tr, id}) continue;
    for (auto x : g[wh]) {
      int nxd, nxt;
      if (x.second <= rem)
        nxd = rem - x.second, nxt = tr;
      else {
        x.second -= rem;
        nxt = tr + (x.second + vel - 1) / vel;
        nxd = (vel - x.second % vel) % vel;
      }
      if (dist[x.first] > (pair<int, int>){nxt, id}) {
        dist[x.first] = {nxt, id};
        q.push({{nxt, id}, {nxd, x.first}});
      }
    }
  }
}
int dfs(int pos, int par = -1) {
  tim[pos] = idd++;
  sz[pos] = 1;
  for (auto x : gr[pos])
    if (x != par) sz[pos] += dfs(x, pos);
  return sz[pos];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1, ThxDem = n; i < ThxDem; ++i) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  dfs(0);
  lca_init();
  int q;
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<pair<int, int>> vir(k);
    vector<int> spe(m);
    for (int i = 0, ThxDem = k; i < ThxDem; ++i) {
      cin >> vir[i].first >> vir[i].second;
      vir[i].first--;
    }
    for (int i = 0, ThxDem = m; i < ThxDem; ++i) cin >> spe[i], spe[i]--;
    go(vir, spe);
    for (auto x : spe) cout << dist[x].second + 1 << " ";
    cout << "\n";
  }
}