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 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
int h[maxn], sp[20][maxn], st[maxn], C = 0;
vector<int> v[maxn];
void dfs(int u, int par = 0) {
  h[u] = h[par] + 1;
  st[u] = C++;
  sp[0][u] = par;
  for (int i = 1; i < 20; i++) sp[i][u] = sp[i - 1][sp[i - 1][u]];
  for (int y : v[u])
    if (y != par) dfs(y, u);
}
int lca(int a, int b) {
  if (h[a] < h[b]) swap(a, b);
  for (int i = 19; i >= 0; i--)
    if (h[sp[i][a]] >= h[b]) a = sp[i][a];
  if (a == b) return a;
  for (int i = 19; i >= 0; i--)
    if (sp[i][a] != sp[i][b]) a = sp[i][a], b = sp[i][b];
  return sp[0][a];
}
int spd[maxn], beg[maxn], imp[maxn];
int col[maxn], STH_STRANGE[maxn], ext[maxn];
bool mark[maxn];
vector<int> verts;
vector<pair<int, int> > g[maxn];
priority_queue<pair<pair<int, int>, int> > pq;
int mkay(int a, int b) { return ((a % b) + b) % b; }
void add(int u) {
  mark[u] = 1;
  for (pair<int, int> p : g[u]) {
    if (mark[p.first]) continue;
    int t =
        STH_STRANGE[u] + ((p.second - ext[u] + spd[col[u]] - 1) / spd[col[u]]);
    if ((pair<int, int>){t, col[u]} <
        (pair<int, int>){STH_STRANGE[p.first], col[p.first]})
      ext[p.first] = mkay(-p.second + ext[u], spd[col[u]]),
      col[p.first] = col[u], STH_STRANGE[p.first] = t,
      pq.push({{-STH_STRANGE[p.first], -col[u]}, p.first});
  }
}
void init() {
  while (int((pq).size())) pq.pop();
  for (int u : verts) col[u] = -1, STH_STRANGE[u] = inf, mark[u] = 0;
}
void solve(int n) {
  init();
  for (int i = 0; i < n; i++) {
    col[beg[i]] = i;
    STH_STRANGE[beg[i]] = 0;
    ext[beg[i]] = 0;
    add(beg[i]);
  }
  while (int((pq).size())) {
    auto p = pq.top();
    pq.pop();
    if (mark[p.second]) continue;
    add(p.second);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie();
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  dfs(1);
  int q;
  cin >> q;
  while (q--) {
    int N, M;
    cin >> N >> M;
    verts.clear();
    for (int i = 0; i < N; i++) {
      cin >> beg[i] >> spd[i];
      verts.push_back(beg[i]);
    }
    for (int i = 0; i < M; i++) {
      cin >> imp[i];
      verts.push_back(imp[i]);
    }
    sort(verts.begin(), verts.end(),
         [&](int i, int j) { return st[i] < st[j]; });
    verts.resize(unique(verts.begin(), verts.end()) - verts.begin());
    for (int i = int((verts).size()) - 1; i > 0; i--)
      verts.push_back(lca(verts[i], verts[i - 1]));
    sort(verts.begin(), verts.end(),
         [&](int i, int j) { return st[i] < st[j]; });
    verts.resize(unique(verts.begin(), verts.end()) - verts.begin());
    for (int u : verts) g[u].clear();
    for (int i = 1; i < int((verts).size()); i++) {
      int A = verts[i], B = lca(verts[i - 1], verts[i]);
      g[A].push_back({B, h[A] - h[B]});
      g[B].push_back({A, h[A] - h[B]});
    }
    solve(N);
    for (int i = 0; i < M; i++) cout << col[imp[i]] + 1 << " ";
    cout << "\n";
  }
  return 0;
}