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;
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (T& x : v) is >> x;
  return is;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  if (!v.empty()) {
    os << v.front();
    for (int i = 1; i < v.size(); ++i) os << ' ' << v[i];
  }
  return os;
}
struct Node {
  pair<int, int> a;
  int b;
  Node(const pair<int, int>& a, int b) : a(a), b(b) {}
  bool operator>(const Node& rhs) const { return a > rhs.a; }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int rep = 1; rep < n; ++rep) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<bool> vis(n);
  vector<int> prt(n, -1), dep(n), sz(n, 1), pref(n, -1);
  function<void(int)> dfs = [&](int u) {
    vis[u] = true;
    for (int v : g[u])
      if (!vis[v]) {
        dep[v] = dep[u] + 1;
        prt[v] = u;
        dfs(v);
        sz[u] += sz[v];
        if (pref[u] == -1 || sz[v] > sz[pref[u]]) pref[u] = v;
      }
  };
  dfs(0);
  vector<int> top(n, -1), dfn(n);
  int t = 0;
  function<void(int)> hld = [&](int u) {
    dfn[u] = t++;
    if (top[u] == -1) top[u] = u;
    if (pref[u] != -1) {
      top[pref[u]] = top[u];
      hld(pref[u]);
      for (int v : g[u])
        if (v != prt[u] && v != pref[u]) hld(v);
    }
  };
  hld(0);
  function<int(int, int)> lca = [&](int u, int v) {
    int x = top[u], y = top[v];
    while (x != y) {
      if (dep[x] > dep[y]) {
        swap(x, y);
        swap(u, v);
      }
      v = prt[y];
      y = top[v];
    }
    return dep[u] < dep[v] ? u : v;
  };
  function<int(int, int)> dist = [&](int u, int v) {
    return dep[u] + dep[v] - (dep[lca(u, v)] << 1);
  };
  int q;
  cin >> q;
  stack<int> stk;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> v(k), s(k), uu(m);
    for (int i = 0; i < k; ++i) cin >> v[i] >> s[i];
    cin >> uu;
    vector<pair<int, int>> ver;
    for (int& x : v) {
      --x;
      ver.emplace_back(dfn[x], x);
    }
    for (int& x : uu) {
      --x;
      ver.emplace_back(dfn[x], x);
    }
    sort(ver.begin(), ver.end());
    ver.erase(unique(ver.begin(), ver.end()), ver.end());
    map<int, int> mp;
    t = 0;
    vector<vector<int>> vir;
    vector<int> vu((k + m) * 2);
    vir.resize((k + m) * 2);
    function<int(int)> id = [&](int u) {
      auto it = mp.find(u);
      if (it != mp.end()) return it->second;
      vu[t] = u;
      mp.insert(make_pair(u, t));
      return t++;
    };
    function<void(int, int)> link = [&](int u, int v) {
      int x = id(u), y = id(v);
      vir[x].push_back(y);
      vir[y].push_back(x);
    };
    for (const auto& pr : ver) {
      int u = pr.second;
      if (stk.empty()) {
        stk.push(u);
        id(u);
        continue;
      }
      int prv = -1;
      int lc = lca(u, stk.top());
      while (!stk.empty() && dep[stk.top()] > dep[lc]) {
        if (prv != -1) link(prv, stk.top());
        prv = stk.top();
        stk.pop();
      }
      if (prv != -1) link(prv, lc);
      if (stk.empty() || stk.top() != lc) stk.push(lc);
      if (stk.top() != u) stk.push(u);
    }
    int prv = -1;
    while (!stk.empty()) {
      if (prv != -1) link(prv, stk.top());
      prv = stk.top();
      stk.pop();
    }
    vector<pair<int, int>> occ(
        t, make_pair(numeric_limits<int>::max(), numeric_limits<int>::max()));
    priority_queue<Node, vector<Node>, greater<Node>> que;
    for (int i = 0; i < k; ++i)
      que.emplace(occ[id(v[i])] = make_pair(0, i), id(v[i]));
    while (!que.empty()) {
      Node tmp = que.top();
      que.pop();
      if (occ[tmp.b] != tmp.a) continue;
      int i = tmp.a.second;
      for (int x : vir[tmp.b]) {
        pair<int, int> di =
            make_pair((dist(v[i], vu[x]) + s[i] - 1) / s[i], tmp.a.second);
        if (di < occ[x]) que.emplace(occ[x] = di, x);
      }
    }
    for (int i = 0; i < m; ++i) cout << (occ[id(uu[i])].second + 1) << ' ';
    cout << '\n';
  }
  return 0;
}