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;
using ll = long long;
using pll = pair<ll, ll>;
const int N = 2e5 + 5;
int n, dep[N], dfn[N], ts, fa[N][20], ans[N];
pair<int, int> turn[N];
vector<int> g[N];
vector<pair<int, int>> vg[N];
void dfs(int u, int p, int d) {
  dep[u] = d, dfn[u] = ++ts;
  fa[u][0] = p;
  for (int i = 1; i < 20; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (int v : g[u]) {
    if (v == p) continue;
    dfs(v, u, d + 1);
  }
}
int lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int d = dep[u] - dep[v];
  for (int i = 0; i < 20; ++i)
    if (d & 1 << i) u = fa[u][i];
  if (u == v) return u;
  for (int i = 19; i >= 0; --i)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
void connect(int u, int v) {
  int dist = abs(dep[u] - dep[v]);
  vg[u].emplace_back(v, dist);
  vg[v].emplace_back(u, dist);
}
void build_vtree(vector<int> key_nodes, int root = 1) {
  assert(dfn[root] == 1);
  sort(key_nodes.begin(), key_nodes.end(),
       [&](int u, int v) { return dfn[u] < dfn[v]; });
  vector<int> st;
  auto ins = [&](int u) {
    st.push_back(u);
    vg[u].clear();
    ans[u] = INT_MAX;
    turn[u] = make_pair(INT_MAX, INT_MAX);
  };
  ins(root);
  for (int u : key_nodes) {
    if (u == root) continue;
    int an = lca(u, st.back());
    if (an != st.back()) {
      vector<int> link;
      while (dfn[st.back()] > dfn[an]) link.push_back(st.back()), st.pop_back();
      reverse(link.begin(), link.end());
      if (st.back() != an) ins(an);
      connect(an, link.front());
      for (int i = 0, sz = link.size(); i + 1 < sz; ++i)
        connect(link[i], link[i + 1]);
    }
    ins(u);
  }
  for (int i = 0, sz = st.size(); i + 1 < sz; ++i) connect(st[i], st[i + 1]);
}
int main() {
  ts = 0;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  dfs(1, 1, 0);
  int q;
  cin >> q;
  while (q--) {
    int k, m;
    scanf("%d%d", &k, &m);
    vector<int> key_nodes, cities;
    key_nodes.push_back(1);
    vector<pair<int, int>> viruses;
    for (int i = 0; i < k; ++i) {
      int v, s;
      scanf("%d%d", &v, &s);
      viruses.emplace_back(v, s);
      key_nodes.push_back(v);
    }
    for (int i = 0; i < m; ++i) {
      int x;
      scanf("%d", &x);
      cities.push_back(x);
      key_nodes.push_back(x);
    }
    sort(key_nodes.begin(), key_nodes.end());
    key_nodes.resize(unique(key_nodes.begin(), key_nodes.end()) -
                     key_nodes.begin());
    build_vtree(key_nodes);
    using State = tuple<int, int, int, int>;
    priority_queue<State, vector<State>, greater<State>> pq;
    for (int i = 0; i < viruses.size(); ++i) {
      auto [v, s] = viruses[i];
      turn[v] = {0, i}, ans[v] = i;
      pq.emplace(0, i, v, 0);
    }
    while (!pq.empty()) {
      auto [t, id, u, r] = pq.top();
      pq.pop();
      if (make_pair(t, id) > turn[u]) continue;
      int speed = viruses[id].second;
      for (auto [v, len] : vg[u]) {
        int nt, nr;
        if (r >= len) {
          nt = t, nr = r - len;
        } else {
          nt = t + (len - r + speed - 1) / speed;
          nr = (nt - t) * speed + r - len;
        }
        pair<int, int> ns = make_pair(nt, id);
        if (ns < turn[v]) {
          turn[v] = ns;
          ans[v] = id;
          pq.emplace(nt, id, v, nr);
        }
      }
    }
    for (int u : cities) printf("%d ", ans[u] + 1);
    puts("");
  }
}