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 N = 200005, LG = 18;
int n, ts, St[N], Fn[N], H[N], P[N][LG];
int q, V[N], S[N], U[N], C[N];
pair<int, int> D[N];
vector<int> Adj[N], Ad[N];
void DFS(int v, int p) {
  St[v] = ts++;
  P[v][0] = p;
  for (int i = 1; i < LG; i++) P[v][i] = P[P[v][i - 1]][i - 1];
  for (int u : Adj[v])
    if (u != p) H[u] = H[v] + 1, DFS(u, v);
  Fn[v] = ts;
}
inline int LCA(int v, int u) {
  if (H[v] < H[u]) return (LCA(u, v));
  for (int i = 0; i < LG; i++)
    if ((H[v] - H[u]) >> i & 1) v = P[v][i];
  if (v == u) return (v);
  for (int i = LG - 1; ~i; i--)
    if (P[v][i] != P[u][i]) v = P[v][i], u = P[u][i];
  return (P[v][0]);
}
inline int Dist(int v, int u) { return (H[v] + H[u] - H[LCA(v, u)] * 2); }
inline bool Parent(int p, int v) { return (St[p] <= St[v] && Fn[v] <= Fn[p]); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int v, u;
    scanf("%d%d", &v, &u);
    Adj[v].push_back(u);
    Adj[u].push_back(v);
  }
  DFS(1, 0);
  scanf("%d", &q);
  for (; q; q--) {
    int k, m;
    vector<int> vec;
    scanf("%d%d", &k, &m);
    for (int i = 1; i <= k; i++)
      scanf("%d%d", &V[i], &S[i]), vec.push_back(V[i]);
    for (int i = 1; i <= m; i++) scanf("%d", &U[i]), vec.push_back(U[i]);
    sort(vec.begin(), vec.end(), [&](int i, int j) { return St[i] < St[j]; });
    vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
    int tmp = (int)vec.size();
    for (int i = 0; i < tmp - 1; i++) vec.push_back(LCA(vec[i], vec[i + 1]));
    sort(vec.begin(), vec.end(), [&](int i, int j) { return St[i] < St[j]; });
    vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
    vector<int> pars = {vec[0]};
    for (int i = 1; i < (int)vec.size(); i++) {
      while (!Parent(pars.back(), vec[i])) pars.pop_back();
      Ad[pars.back()].push_back(vec[i]);
      Ad[vec[i]].push_back(pars.back());
      pars.push_back(vec[i]);
    }
    priority_queue<pair<pair<int, int>, int> > Pq;
    for (int v : vec) D[v] = {N, N};
    for (int i = 1; i <= k; i++) D[V[i]] = {0, i}, Pq.push({{0, -i}, V[i]});
    while (Pq.size()) {
      pair<int, int> d = Pq.top().first;
      int v = Pq.top().second;
      Pq.pop();
      d.first = -d.first;
      d.second = -d.second;
      if (d > D[v]) continue;
      int c = d.second;
      C[v] = c;
      for (int u : Ad[v]) {
        pair<int, int> du = {(Dist(V[c], u) + S[c] - 1) / S[c], c};
        if (du >= D[u]) continue;
        D[u] = du;
        du.first = -du.first;
        du.second = -du.second;
        Pq.push({du, u});
      }
    }
    for (int i = 1; i <= m; i++) printf("%d ", C[U[i]]);
    printf("\n");
    for (int v : vec) Ad[v].clear();
  }
  return 0;
}