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>
const int P = 998244353;
const int B = 19;
const int N = 1 << B;
const int SZ = 202020;
using namespace std;
int n, q, dfso[SZ], lvl[SZ];
pair<int, int> etl[2 * N];
vector<int> g[SZ];
vector<pair<int, int> > c[SZ];
int fe[SZ];
int te = 0;
void dfs(int i, int l) {
  lvl[i] = l;
  dfso[i] = ++te;
  etl[te + N] = {l, i};
  for (int j : g[i]) {
    if (!lvl[j]) {
      dfs(j, l + 1);
      te++;
      etl[te + N] = {l, i};
    }
  }
}
void mklca() {
  for (int i = 1; i < 2 * N; i++) etl[i] = {1e9, 1e9};
  dfs(1, 1);
  for (int i = N - 1; i > 0; i--) etl[i] = min(etl[2 * i], etl[2 * i + 1]);
}
int glca(int a, int b) {
  a = dfso[a] + N;
  b = dfso[b] + N;
  if (a > b) swap(a, b);
  pair<int, int> x = etl[a];
  while (a <= b) {
    if (a & 1) x = min(x, etl[a++]);
    if (~b & 1) x = min(x, etl[b--]);
    a /= 2;
    b /= 2;
  }
  return x.second;
}
bool cmp(int &a, int &b) { return dfso[a] < dfso[b]; }
vector<int> vtx;
vector<pair<int, int> > edg;
void insubt(vector<int> &nds) {
  sort(nds.begin(), nds.end(), cmp);
  vtx = nds;
  int nz = nds.size();
  vector<int> adj(nz), plca(nz);
  adj[0] = -1;
  for (int i = 1; i < nz; i++) {
    plca[i] = glca(nds[i - 1], nds[i]);
    adj[i] = i - 1;
  }
  int po = 1;
  for (int i = 0; i < nz - 2; i++) {
    while (po + 1 < nz && lvl[plca[po]] < lvl[plca[po + 1]]) po++;
    int nlca = glca(nds[po], nds[adj[po]]);
    bool k1 = (nds[po] != nlca), k2 = (nds[adj[po]] != nlca);
    if (k1) edg.push_back({nlca, nds[po]});
    if (k2) edg.push_back({nlca, nds[adj[po]]});
    if (k1 && k2) vtx.push_back(nlca);
    nds[po] = nlca;
    adj[po] = adj[adj[po]];
    if (adj[po] >= 0) plca[po] = glca(nds[po], nds[adj[po]]);
    if (po + 1 < nz) plca[po + 1] = glca(nds[po], nds[po + 1]);
    if (adj[po] < 0) po++;
  }
  if (nz > 1) edg.push_back({nds[po], nds[adj[po]]});
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  mklca();
  cin >> q;
  for (; q--;) {
    int k, m;
    cin >> k >> m;
    vector<int> imp(m), dc;
    vector<pair<int, int> > vr(k);
    for (int i = 0; i < k; i++) {
      cin >> vr[i].first >> vr[i].second;
      dc.push_back(vr[i].first);
    }
    for (int i = 0; i < m; i++) {
      cin >> imp[i];
      dc.push_back(imp[i]);
    }
    edg.clear();
    insubt(dc);
    for (pair<int, int> p : edg) {
      int dist =
          lvl[p.first] + lvl[p.second] - 2 * lvl[glca(p.first, p.second)];
      c[p.first].push_back({p.second, dist});
      c[p.second].push_back({p.first, dist});
    }
    priority_queue<pair<pair<int, int>, pair<int, int> > > qd;
    for (int i = 0; i < k; i++) {
      fe[vr[i].first] = i + 1;
      qd.push({{0, -i}, {vr[i].second, vr[i].first}});
    }
    while (!qd.empty()) {
      pair<pair<int, int>, pair<int, int> > cr = qd.top();
      qd.pop();
      cr.first.second = -cr.first.second;
      if (fe[cr.second.second] && fe[cr.second.second] != cr.first.second + 1)
        continue;
      fe[cr.second.second] = cr.first.second + 1;
      for (pair<int, int> j : c[cr.second.second]) {
        if (!fe[j.first]) {
          pair<pair<int, int>, pair<int, int> > nx = cr;
          nx.second.first -= j.second;
          if (nx.second.first < 0) {
            long long k = 0;
            for (long long b = 1 << 18; b > 0; b /= 2) {
              if ((long long)nx.second.first +
                      (k + b) * (long long)vr[cr.first.second].second <
                  0)
                k += b;
            }
            k++;
            nx.second.first += k * vr[cr.first.second].second;
            nx.first.first -= k;
          }
          nx.first.second = -nx.first.second;
          nx.second.second = j.first;
          qd.push(nx);
        }
      }
    }
    for (int i : imp) cout << fe[i] << ' ';
    cout << '\n';
    for (int i : vtx) {
      c[i].clear();
      fe[i] = 0;
    }
  }
  return 0;
}