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>
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
class HLD {
 public:
  void init(int n) {
    sz.resize(n);
    in.resize(n);
    out.resize(n);
    rin.resize(n);
    p.resize(n);
    edges.resize(n);
    nxt.resize(n);
    h.resize(n);
  }
  void addEdge(int u, int v) {
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  void setRoot(int n) {
    t = 0;
    p[n] = n;
    h[n] = 0;
    prep(n, n);
    nxt[n] = n;
    hld(n);
  }
  int getLCA(int u, int v) {
    while (!inSubtree(nxt[u], v)) {
      u = p[nxt[u]];
    }
    while (!inSubtree(nxt[v], u)) {
      v = p[nxt[v]];
    }
    return in[u] < in[v] ? u : v;
  }
  bool inSubtree(int u, int v) { return in[u] <= in[v] && in[v] < out[u]; }
  std::vector<std::pair<int, int>> getPathtoAncestor(int u, int anc) {
    std::vector<std::pair<int, int>> ans;
    while (nxt[u] != nxt[anc]) {
      ans.emplace_back(in[nxt[u]], in[u] + 1);
      u = p[nxt[u]];
    }
    ans.emplace_back(in[anc], in[u] + 1);
    return ans;
  }
  std::vector<int> in, out, p, rin, sz, nxt, h;
  std::vector<std::vector<int>> edges;
  int t;
  void prep(int on, int par) {
    sz[on] = 1;
    p[on] = par;
    for (int i = 0; i < (int)edges[on].size(); i++) {
      int &u = edges[on][i];
      if (u == par) {
        std::swap(u, edges[on].back());
        edges[on].pop_back();
        i--;
      } else {
        h[u] = 1 + h[on];
        prep(u, on);
        sz[on] += sz[u];
        if (sz[u] > sz[edges[on][0]]) {
          std::swap(edges[on][0], u);
        }
      }
    }
  }
  void hld(int on) {
    in[on] = t++;
    rin[in[on]] = on;
    for (auto u : edges[on]) {
      nxt[u] = (u == edges[on][0] ? nxt[on] : u);
      hld(u);
    }
    out[on] = t;
  }
};
const int INF = 1e9;
const int ms = 200200;
HLD hld;
bool cmp(int u, int v) { return hld.in[u] < hld.in[v]; }
std::vector<int> virtualTree;
std::vector<std::pair<int, int>> edges[ms];
bool visit[ms];
int speed[ms];
int ans[ms];
void solve() {
  int virus, k;
  std::cin >> virus >> k;
  std::vector<int> special(k);
  std::priority_queue<
      std::pair<std::pair<int, int>, std::pair<int, int>>,
      std::vector<std::pair<std::pair<int, int>, std::pair<int, int>>>,
      std::greater<std::pair<std::pair<int, int>, std::pair<int, int>>>>
      hp;
  for (int i = 0; i < virus; i++) {
    int city;
    std::cin >> city >> speed[i];
    city--;
    virtualTree.push_back(city);
    hp.push(std::pair<std::pair<int, int>, std::pair<int, int>>(
        std::pair<int, int>(0, i), std::pair<int, int>(0, city)));
  }
  for (int i = 0; i < k; i++) {
    std::cin >> special[i];
    special[i]--;
    virtualTree.push_back(special[i]);
  }
  std::sort(virtualTree.begin(), virtualTree.end(), cmp);
  for (int i = 0, n = (int)virtualTree.size(); i + 1 < n; i++) {
    virtualTree.push_back(hld.getLCA(virtualTree[i], virtualTree[i + 1]));
  }
  std::sort(virtualTree.begin(), virtualTree.end(), cmp);
  virtualTree.resize(std::unique(virtualTree.begin(), virtualTree.end()) -
                     virtualTree.begin());
  std::stack<int> st;
  for (auto v : virtualTree) {
    if (st.empty()) {
      st.push(v);
    } else {
      while (!st.empty() && !hld.inSubtree(st.top(), v)) st.pop();
      assert(!st.empty());
      edges[st.top()].emplace_back(v, hld.h[v] - hld.h[st.top()]);
      edges[v].emplace_back(st.top(), hld.h[v] - hld.h[st.top()]);
      st.push(v);
    }
  }
  while (!hp.empty()) {
    int on = hp.top().second.second;
    int dist = hp.top().second.first;
    int curVirus = hp.top().first.second;
    hp.pop();
    if (visit[on]) continue;
    visit[on] = true;
    ans[on] = curVirus + 1;
    int s = speed[curVirus];
    for (auto e : edges[on]) {
      int toDist = e.second + dist;
      hp.push(std::pair<std::pair<int, int>, std::pair<int, int>>(
          std::pair<int, int>((toDist + s - 1) / s, curVirus),
          std::pair<int, int>(toDist, e.first)));
    }
  }
  for (int i = 0; i < k; i++) {
    std::cout << ans[special[i]] << (i + 1 == k ? '\n' : ' ');
  }
  for (auto v : virtualTree) {
    assert(visit[v]);
    visit[v] = false;
    edges[v].clear();
  }
  virtualTree.clear();
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  std::cin >> n;
  hld.init(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    std::cin >> u >> v;
    u--;
    v--;
    hld.addEdge(u, v);
  }
  hld.setRoot(0);
  int q;
  std::cin >> q;
  while (q--) {
    solve();
  }
}