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<int> adjList[200000];
int parent[200000][19], depth[200000], disc[200000], fin[200000], num = 0;
int doDFS(int u, int p, int d) {
  int i;
  parent[u][0] = p, depth[u] = d, disc[u] = num++;
  for (i = 0; i < adjList[u].size(); i++) {
    int v = adjList[u][i];
    if (v != p) doDFS(v, u, d + 1);
  }
  fin[u] = num;
  return 0;
}
int lca(int u, int v, int logn) {
  int i;
  if (depth[u] < depth[v]) swap(u, v);
  for (i = logn - 1; i >= 0; i--) {
    if ((parent[u][i] != -1) && (depth[parent[u][i]] >= depth[v]))
      u = parent[u][i];
  }
  if (u == v) return u;
  for (i = logn - 1; i >= 0; i--) {
    if (parent[u][i] != parent[v][i]) u = parent[u][i], v = parent[v][i];
  }
  return parent[u][0];
}
int v[200000], s[200000], u[200000];
vector<pair<int, int> > nodes;
vector<int> S;
vector<pair<int, int> > adjList2[200000];
int dist[200000], virus[200000];
priority_queue<pair<pair<int, int>, int> > H;
int main() {
  int i, j, l;
  int n, q, k, m;
  int x, y;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d", &x, &y);
    x--, y--;
    adjList[x].push_back(y);
    adjList[y].push_back(x);
  }
  doDFS(0, -1, 0);
  for (i = 1; (1 << i) < n; i++) {
    for (j = 0; j < n; j++) {
      if (parent[j][i - 1] != -1)
        parent[j][i] = parent[parent[j][i - 1]][i - 1];
      else
        parent[j][i] = -1;
    }
  }
  int logn = i;
  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    scanf("%d %d", &k, &m);
    for (j = 0; j < k; j++)
      scanf("%d %d", &v[j], &s[j]), v[j]--,
          nodes.push_back(make_pair(disc[v[j]], v[j]));
    for (j = 0; j < m; j++)
      scanf("%d", &u[j]), u[j]--, nodes.push_back(make_pair(disc[u[j]], u[j]));
    sort(nodes.begin(), nodes.end());
    for (j = 1; j < k + m; j++) {
      int l = lca(nodes[j - 1].second, nodes[j].second, logn);
      nodes.push_back(make_pair(disc[l], l));
    }
    sort(nodes.begin(), nodes.end());
    nodes.resize(unique(nodes.begin(), nodes.end()) - nodes.begin());
    for (j = nodes.size() - 1; j >= 0; j--) {
      while (!S.empty() && (disc[S.back()] < fin[nodes[j].second])) {
        adjList2[S.back()].push_back(make_pair(
            nodes[j].second, depth[S.back()] - depth[nodes[j].second]));
        adjList2[nodes[j].second].push_back(
            make_pair(S.back(), depth[S.back()] - depth[nodes[j].second]));
        S.pop_back();
      }
      S.push_back(nodes[j].second);
    }
    for (j = 0; j < nodes.size(); j++)
      dist[nodes[j].second] = virus[nodes[j].second] = -1;
    for (j = 0; j < k; j++)
      dist[v[j]] = 0, virus[v[j]] = j,
      H.push(make_pair(make_pair(0, -j), v[j]));
    while (!H.empty()) {
      int u = H.top().second;
      pair<int, int> d = H.top().first;
      H.pop();
      if (d <
          make_pair(-((dist[u] + s[virus[u]] - 1) / s[virus[u]]), -virus[u]))
        continue;
      for (l = 0; l < adjList2[u].size(); l++) {
        int v = adjList2[u][l].first;
        if (virus[u] == virus[v]) continue;
        int t =
            (dist[u] + adjList2[u][l].second + s[virus[u]] - 1) / s[virus[u]];
        if ((dist[v] == -1) ||
            (make_pair(t, virus[u]) <
             make_pair((dist[v] + s[virus[v]] - 1) / s[virus[v]], virus[v]))) {
          dist[v] = dist[u] + adjList2[u][l].second;
          virus[v] = virus[u];
          H.push(make_pair(make_pair(-t, -virus[v]), v));
        }
      }
    }
    for (j = 0; j < m; j++) printf("%d ", virus[u[j]] + 1);
    printf("\n");
    for (j = 0; j < nodes.size(); j++) adjList2[nodes[j].second].clear();
    S.clear(), nodes.clear();
  }
  return 0;
}