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, inf = 1e9 + 7;
int n, q, k, m, par[maxn][20], st[maxn], fn[maxn], h[maxn], timee, dp[maxn],
    dis[maxn], imp[maxn], root, ar[maxn];
vector<int> graph[maxn], vec;
vector<pair<int, int>> auxgraph[maxn];
bool comp(int a, int b) { return st[a] < st[b]; }
int dfs(int i, int p) {
  st[i] = fn[i] = timee++;
  par[i][0] = p;
  for (int j = 1; j < 20; j++) par[i][j] = par[par[i][j - 1]][j - 1];
  for (int u : graph[i])
    if (u != p) h[u] = h[i] + 1, fn[i] = dfs(u, i);
  return fn[i];
}
int calc(int i, int p) {
  for (int j = 0; j < 20; j++)
    if (p & (1 << j)) i = par[i][j];
  return i;
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  u = calc(u, h[u] - h[v]);
  if (u == v) return u;
  for (int i = 19; i > -1; i--)
    if (par[u][i] != par[v][i]) u = par[u][i], v = par[v][i];
  return par[u][0];
}
void auxiliary() {
  sort(vec.begin(), vec.end(), comp);
  for (int i = 0; i < k + m - 1; i++) vec.push_back(lca(vec[i], vec[i + 1]));
  sort(vec.begin(), vec.end(), comp);
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  stack<int> s;
  root = *vec.begin();
  s.push(root);
  for (int i = 1; i < vec.size(); i++) {
    while (st[s.top()] > st[vec[i]] || fn[s.top()] < st[vec[i]]) s.pop();
    auxgraph[s.top()].push_back({vec[i], h[vec[i]] - h[s.top()]});
    s.push(vec[i]);
  }
}
void update(int i, int j, int w) {
  if (dp[j] == 0) return;
  int ds = dis[j] + w;
  if (dp[i] == 0) dp[i] = dp[j], dis[i] = ds;
  if ((ds + ar[dp[j]] - 1) / ar[dp[j]] < (dis[i] + ar[dp[i]] - 1) / ar[dp[i]])
    dp[i] = dp[j], dis[i] = ds;
  else if ((ds + ar[dp[j]] - 1) / ar[dp[j]] ==
               (dis[i] + ar[dp[i]] - 1) / ar[dp[i]] &&
           dp[j] < dp[i])
    dp[i] = dp[j], dis[i] = ds;
}
void dpdown(int i) {
  for (auto u : auxgraph[i]) {
    dpdown(u.first);
    update(i, u.first, u.second);
  }
}
void dpup(int i) {
  for (auto u : auxgraph[i]) {
    update(u.first, i, u.second);
    dpup(u.first);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  dfs(1, 0);
  cin >> q;
  while (q--) {
    cin >> k >> m;
    for (int i = 1, u, w; i <= k; i++) {
      cin >> u >> w;
      vec.push_back(u);
      dp[u] = i;
      dis[u] = 0;
      ar[i] = w;
    }
    for (int i = 0, u; i < m; i++) {
      cin >> u;
      vec.push_back(u);
      imp[i] = u;
    }
    auxiliary();
    dpdown(root);
    dpup(root);
    for (int i = 0; i < m; i++) cout << dp[imp[i]] << " ";
    cout << endl;
    for (int i : vec) dp[i] = 0, auxgraph[i].clear();
    vec.clear();
  }
}