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 ll = long long;
using namespace std;
const int N = 4e5 + 100;
const int LOG = 18;
int n, in[N], out[N], siz[N], dep[N], jj[N][LOG];
vector<int> G[N];
int dfs(int v, int p) {
  jj[v][0] = p;
  for (int i = 1; i <= LOG - 1; ++i) jj[v][i] = jj[jj[v][i - 1]][i - 1];
  in[v] = ++*in;
  dep[v] = dep[p] + 1;
  siz[v] = 1;
  for (auto u : G[v])
    if (u != p) siz[v] += dfs(u, v);
  out[v] = *in;
  return siz[v];
}
int ispar(int a, int b) { return in[a] <= in[b] && in[b] <= out[a]; }
int lca(int a, int b) {
  for (int i = LOG - 1; 0 <= i; --i)
    if (!ispar(jj[a][i], b)) a = jj[a][i];
  return jj[a][0];
}
int spe[N], sta[N], qw[N], nodes[N], vis[N];
vector<int> graf[N];
struct gao {
  int v, id, czas;
  bool operator<(gao p) const {
    auto sufit = [](int a, int b) { return (a + b - 1) / b; };
    int a = sufit(czas, spe[id]), b = sufit(p.czas, spe[p.id]);
    if (a != b) return a > b;
    return id > p.id;
  }
};
gao dd[N];
void solve() {
  *nodes = 0;
  int k, m;
  scanf("%d%d", &k, &m);
  for (int i = 1; i <= k; ++i) {
    scanf("%d%d", sta + i, spe + i);
    nodes[++*nodes] = sta[i];
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", qw + i);
    nodes[++*nodes] = qw[i];
  }
  sort(nodes + 1, nodes + *nodes + 1,
       [&](int a, int b) { return in[a] < in[b]; });
  *nodes = unique(nodes + 1, nodes + *nodes + 1) - nodes - 1;
  int sz = *nodes;
  for (int i = 1; i <= sz - 1; ++i)
    nodes[++*nodes] = lca(nodes[i], nodes[i + 1]);
  sort(nodes + 1, nodes + *nodes + 1,
       [&](int a, int b) { return in[a] < in[b]; });
  *nodes = unique(nodes + 1, nodes + *nodes + 1) - nodes - 1;
  vector<int> s = {nodes[1]};
  for (int i = 2; i <= *nodes; ++i) {
    while (!ispar(s.back(), nodes[i])) s.pop_back();
    graf[s.back()].push_back(nodes[i]);
    graf[nodes[i]].push_back(s.back());
    s.push_back(nodes[i]);
  }
  priority_queue<gao> qq;
  for (int i = 1; i <= *nodes; ++i) dd[nodes[i]] = {0, 0, (int)1e9};
  for (int i = 1; i <= k; ++i) {
    dd[sta[i]] = {sta[i], i, 0};
    qq.push(dd[sta[i]]);
  }
  while (!qq.empty()) {
    int v = qq.top().v;
    qq.pop();
    if (vis[v]) continue;
    vis[v] = 1;
    for (auto u : graf[v]) {
      int dis = abs(dep[v] - dep[u]) + dd[v].czas;
      gao nw = {u, dd[v].id, dis};
      if (dd[u] < nw) {
        dd[u] = nw;
        qq.push(nw);
      }
    }
  }
  for (int i = 1; i <= m; ++i) printf("%d ", dd[qw[i]].id);
  printf("\n");
  for (int i = 1; i <= *nodes; ++i) {
    vis[nodes[i]] = 0;
    graf[nodes[i]].clear();
  }
}
int main() {
  spe[0] = 1;
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 1);
  int q;
  scanf("%d", &q);
  while (q--) solve();
  return 0;
}