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;
int n, m, k, qq, h[200005], par[200005][18], at[200005], s[200005], u[200005],
    cnt, ord[200005];
vector<int> v[200005], vv[200005];
pair<int, int> p[200005];
struct cmp {
  bool operator()(const pair<pair<int, int>, int>& a,
                  const pair<pair<int, int>, int>& b) {
    int ta = a.first.first / s[a.first.second];
    int tb = b.first.first / s[b.first.second];
    if (ta != tb) return ta > tb;
    if (a.first.second != b.first.second)
      return a.first.second > b.first.second;
    return a.second > b.second;
  }
  bool operator()(const pair<int, int>& a, const pair<int, int>& b) {
    int ta = a.first / s[a.second];
    int tb = b.first / s[b.second];
    if (ta != tb) return ta > tb;
    return a.second > b.second;
  }
};
priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int>>,
               cmp>
    q;
void dfs(int a, int b) {
  par[a][0] = b;
  for (int i = 1; i < 18; i++) par[a][i] = par[par[a][i - 1]][i - 1];
  for (int i : v[a])
    if (i != b) {
      h[i] = h[a] + 1;
      dfs(i, a);
    }
  ord[a] = cnt++;
}
int lca(int a, int b) {
  if (h[a] > h[b]) swap(a, b);
  for (int i = 0; i < 18; i++)
    if ((h[b] - h[a]) & (1 << i)) b = par[b][i];
  if (a == b) return a;
  for (int i = 17; i >= 0; i--)
    if (par[a][i] != par[b][i]) {
      a = par[a][i];
      b = par[b][i];
    }
  return par[a][0];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int ta, tb;
    scanf("%d%d", &ta, &tb);
    v[ta].push_back(tb);
    v[tb].push_back(ta);
  }
  dfs(1, 0);
  scanf("%d", &qq);
  while (qq--) {
    vector<int> ss;
    scanf("%d%d", &k, &m);
    for (int i = 0; i < k; i++) {
      scanf("%d%d", at + i, s + i);
      ss.push_back(at[i]);
    }
    for (int i = 0; i < m; i++) {
      scanf("%d", u + i);
      ss.push_back(u[i]);
    }
    sort(ss.begin(), ss.end(), [&](int a, int b) { return ord[a] < ord[b]; });
    if (ss.back() != 1) ss.push_back(1);
    vector<int> st{1}, lc;
    for (int i : ss) {
      int tmp = lca(i, st.back());
      while (h[tmp] < h[st.back()] && h[tmp] <= h[st[st.size() - 2]]) {
        int wtf = st.back();
        st.pop_back();
        vv[wtf].push_back(st.back());
        vv[st.back()].push_back(wtf);
      }
      if (h[tmp] > h[st.back()])
        st.push_back(tmp);
      else if (h[tmp] < h[st.back()]) {
        int wtf = st.back();
        st.pop_back();
        vv[wtf].push_back(tmp);
        vv[tmp].push_back(wtf);
        st.push_back(tmp);
      }
      if (i != st.back()) st.push_back(i);
      p[i].first = INT_MAX;
      p[tmp].first = INT_MAX;
      lc.push_back(tmp);
    }
    for (int i = 0; i < k; i++) q.emplace(p[at[i]] = {s[i] - 1, i}, at[i]);
    while (!q.empty()) {
      auto ta = q.top().first;
      int tb = q.top().second;
      q.pop();
      if (p[tb] != ta) continue;
      for (int i : vv[tb]) {
        auto tc = make_pair(ta.first + abs(h[tb] - h[i]), ta.second);
        if (cmp{}(p[i], tc)) q.emplace(p[i] = tc, i);
      }
    }
    for (int i = 0; i < m; i++)
      printf("%d%c", p[u[i]].second + 1, " \n"[i == m - 1]);
    for (int i : ss) vector<int>().swap(vv[i]);
    for (int i : lc) vector<int>().swap(vv[i]);
  }
}