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;
vector<vector<int> > g;
vector<int> tin;
vector<int> tout;
int timer = 0;
vector<vector<int> > up;
vector<set<pair<int, int> > > t;
vector<int> start;
vector<int> speed;
vector<int> imp;
vector<int> p;
vector<int> h;
vector<pair<int, int> > dp;
set<pair<int, int> > lst;
int logn;
int n;
int inf = 1e9 + 7;
void dfs(int v, int prev) {
  h[v] = h[prev] + 1;
  tin[v] = timer++;
  up[v][0] = prev;
  for (int i = 1; i < logn; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
  }
  for (auto u : g[v]) {
    if (u != prev) dfs(u, v);
  }
  tout[v] = timer++;
}
inline bool check(int u, int v) {
  return tin[u] >= tin[v] && tout[u] <= tout[v];
}
inline int lca(int u, int v) {
  if (check(u, v)) return v;
  if (check(v, u)) return u;
  for (int i = logn - 1; i >= 0; --i) {
    if (!check(u, up[v][i])) v = up[v][i];
  }
  return up[v][0];
}
inline int dist(int u, int v) { return h[u] + h[v] - 2 * h[lca(u, v)]; }
inline void input(int k, int m) {
  start.assign(k, 0);
  speed.assign(k, 0);
  imp.assign(m, 0);
  for (int i = 0; i < k; ++i) {
    cin >> start[i] >> speed[i];
    --start[i];
  }
  for (int i = 0; i < m; ++i) {
    cin >> imp[i];
    --imp[i];
  }
}
inline void build() {
  lst.clear();
  for (auto v : start) lst.insert({tin[v], v});
  for (auto v : imp) lst.insert({tin[v], v});
  int cur = -1;
  for (auto v : lst) {
    if (cur == -1) {
      cur = v.second;
      continue;
    }
    int l = lca(v.second, cur);
    lst.insert({tin[l], l});
    cur = v.second;
  }
  vector<int> st;
  for (auto u : lst) {
    int v = u.second;
    if (st.empty()) {
      st.push_back(v);
      continue;
    }
    while (!st.empty() && !check(v, st.back())) st.pop_back();
    if (st.empty()) {
      st.push_back(v);
      continue;
    }
    int len = dist(st.back(), v);
    t[st.back()].insert({v, len});
    t[v].insert({st.back(), len});
    st.push_back(v);
  }
}
inline void destroy() {
  for (auto v : lst) {
    t[v.second].clear();
    dp[v.second] = make_pair(inf, -1);
  }
}
void dfs1(int v, int prev) {
  for (auto u : t[v]) {
    int to = u.first;
    int len = u.second;
    if (to != prev)
      dfs1(to, v);
    else
      continue;
    if (dp[to].second >= 0)
      dp[v] = min(dp[v], make_pair((dist(v, start[dp[to].second]) +
                                    speed[dp[to].second] - 1) /
                                       speed[dp[to].second],
                                   dp[to].second));
  }
}
void dfs2(int v, int prev) {
  if (prev != -1) {
    if (dp[prev].second >= 0)
      dp[v] = min(dp[v], make_pair((dist(v, start[dp[prev].second]) +
                                    speed[dp[prev].second] - 1) /
                                       speed[dp[prev].second],
                                   dp[prev].second));
  }
  for (auto u : t[v]) {
    int to = u.first;
    if (to != prev) dfs2(to, v);
  }
}
inline void solve() {
  int sz = start.size();
  for (int i = 0; i < sz; ++i) {
    dp[start[i]] = {0, i};
  }
  auto y = *(lst.begin());
  dfs1(y.second, -1);
  dfs2(y.second, -1);
}
inline void output() {
  for (auto v : imp) {
    cout << dp[v].second + 1 << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  g.resize(n);
  tin.resize(n);
  tout.resize(n);
  h.resize(n);
  t.resize(n);
  p.assign(n, 0);
  dp.assign(n, make_pair(inf, -1));
  logn = 0;
  int nn = n;
  while (nn) {
    ++logn;
    nn >>= 1;
  }
  up.assign(n, vector<int>(logn, 0));
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int q;
  cin >> q;
  h[0] = -1;
  dfs(0, 0);
  while (q--) {
    int k, m;
    cin >> k >> m;
    input(k, m);
    build();
    solve();
    output();
    destroy();
  }
  return 0;
}