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 N = 2e5 + 10, K = 20;
vector<int> g[N];
vector<pair<int, int>> ng[N];
int p[N][K];
int tin[N], tout[N], t;
int d[N];
int who[N];
void dfs(int v = 0) {
  tin[v] = t++;
  for (int i = 1; i < K; ++i) {
    p[v][i] = p[p[v][i - 1]][i - 1];
  }
  for (int to : g[v]) {
    if (to != p[v][0]) {
      p[to][0] = v;
      d[to] = d[v] + 1;
      dfs(to);
    }
  }
  tout[v] = t++;
}
bool anc(int u, int v) { return tin[u] <= tin[v] and tout[u] >= tout[v]; }
int lca(int u, int v) {
  if (anc(u, v)) {
    return u;
  }
  if (anc(v, u)) {
    return v;
  }
  for (int i = K - 1; i >= 0; --i) {
    if (!anc(p[u][i], v)) {
      u = p[u][i];
    }
  }
  return p[u][0];
}
void norm(vector<int>& a) {
  sort(a.begin(), a.end(), [&](int u, int v) { return tin[u] < tin[v]; });
  a.erase(unique(a.begin(), a.end()), a.end());
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back(v - 1), g[v - 1].push_back(u - 1);
  }
  dfs();
  int q;
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> s(k), vir(k), vt, sp(m);
    for (int i = 0; i < k; ++i) {
      cin >> vir[i] >> s[i];
      --vir[i];
      vt.push_back(vir[i]);
    }
    for (int i = 0; i < m; ++i) {
      cin >> sp[i];
      --sp[i];
      vt.push_back(sp[i]);
    }
    norm(vt);
    int sz = vt.size();
    for (int i = 0; i + 1 < sz; ++i) {
      vt.push_back(lca(vt[i], vt[i + 1]));
    }
    norm(vt);
    stack<int> st;
    sz = vt.size();
    for (int i = 0; i < sz; ++i) {
      ng[vt[i]].clear();
      who[vt[i]] = -1;
    }
    for (int i = 0; i < sz; ++i) {
      int v = vt[i];
      while (!st.empty() and !anc(st.top(), v)) {
        st.pop();
      }
      if (!st.empty()) {
        int u = st.top();
        int dist = d[v] - d[u];
        assert(dist > 0);
        ng[v].emplace_back(u, dist);
        ng[u].emplace_back(v, dist);
      }
      st.push(v);
    }
    set<tuple<int, int, int, int>> pq;
    auto go = [&](int v, int dist) {
      for (auto [to, dd] : ng[v]) {
        if (who[to] != -1) {
          continue;
        }
        pq.emplace((dd + dist + s[who[v]] - 1) / s[who[v]], who[v], dd + dist,
                   to);
      }
    };
    for (int i = 0; i < k; ++i) {
      who[vir[i]] = i;
    }
    for (int i = 0; i < k; ++i) {
      go(vir[i], 0);
    }
    while (!pq.empty()) {
      auto [D, id, dd, v] = *pq.begin();
      pq.erase(pq.begin());
      if (who[v] != -1) {
        continue;
      }
      who[v] = id;
      go(v, dd);
    }
    for (int i = 0; i < m; ++i) {
      cout << who[sp[i]] + 1 << ' ';
    }
    cout << '\n';
  }
  return 0;
}