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;
const int maxn = 2e5 + 20, maxb = 22, inf = 1e9 + 7;
int n, q, k, m, par[maxn][maxb], st[maxn], en[maxn], h[maxn], tim, dp[maxn],
    dis[maxn], root, speed[maxn], infet[maxn];
vector<int> adj[maxn], ans;
vector<pair<int, int>> adjcm[maxn], vec;
void dfs(int v, int p) {
  h[v] = h[p] + 1;
  par[v][0] = p;
  st[v] = tim++;
  for (int i = 1; i < maxb; i++) par[v][i] = par[par[v][i - 1]][i - 1];
  for (int u : adj[v])
    if (u != p) dfs(u, v);
  en[v] = tim++;
}
int f_p(int v, int bi) {
  for (int i = 0; i < maxb; i++)
    if ((1 << i) & bi) v = par[v][i];
  return v;
}
int lca(int v, int u) {
  if (h[v] < h[u]) swap(v, u);
  if (u == (v = f_p(v, h[v] - h[u]))) return v;
  for (int i = maxb - 1; i >= 0; i--)
    if (par[v][i] != par[u][i]) v = par[v][i], u = par[u][i];
  return par[v][0];
}
void update(int i, int j, int w) {
  if (infet[j] == inf) return;
  int ds = dis[j] + w;
  int tmpinfet = (ds + speed[dp[j]] - 1) / speed[dp[j]];
  if (infet[i] > tmpinfet || (tmpinfet == infet[i] && dp[j] < dp[i])) {
    infet[i] = tmpinfet;
    dis[i] = ds;
    dp[i] = dp[j];
  }
}
void dp_d(int v) {
  for (auto [u, w] : adjcm[v]) {
    dp_d(u);
    update(v, u, w);
  }
}
void dp_u(int v) {
  for (auto [u, w] : adjcm[v]) {
    update(u, v, w);
    dp_u(u);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1, v, u; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  cin >> q;
  for (int i = 0; i < maxn; i++) dp[i] = infet[i] = inf;
  while (q--) {
    cin >> k >> m;
    for (int i = 1, v, w; i <= k; i++) {
      cin >> v >> w;
      vec.push_back({st[v], v});
      dp[v] = i;
      infet[v] = dis[v] = 0;
      speed[i] = w;
    }
    for (int i = 0, v; i < m; i++) {
      cin >> v;
      vec.push_back({st[v], v});
      ans.push_back(v);
    }
    sort(vec.begin(), vec.end());
    for (int i = 0, tmp; i < k + m - 1; i++)
      vec.push_back({st[tmp = lca(vec[i].second, vec[i + 1].second)], tmp});
    sort(vec.begin(), vec.end());
    vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
    stack<int> sk;
    sk.push(root = vec[0].second);
    for (int i = 1; i < vec.size(); i++) {
      while (en[sk.top()] < vec[i].first) sk.pop();
      adjcm[sk.top()].push_back(
          {vec[i].second, h[vec[i].second] - h[sk.top()]});
      sk.push(vec[i].second);
    }
    dp_d(root), dp_u(root);
    for (int i : ans) cout << dp[i] << " ";
    cout << endl;
    for (pair<int, int> i : vec)
      infet[i.second] = dp[i.second] = inf, adjcm[i.second].clear();
    vec.clear(), ans.clear();
  }
  return 0;
}