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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 200010, LOG = 18;
int n, q, m, k, u, v, x, y, t, a, b, ans;
int A[MAXN * 4], N;
int stt[MAXN], fnt[MAXN], timer = 1;
int par[LOG][MAXN], h[MAXN];
int QQQ[MAXN], V[MAXN], S[MAXN], limit[MAXN];
int dist[MAXN], dist2[MAXN];
bool mark[MAXN];
int color[MAXN];
vector<int> G[MAXN], vec[MAXN];
vector<pair<int, int>> G2[MAXN];
void dfs1(int node) {
  stt[node] = timer++;
  for (int i = 1; i < LOG; i++) par[i][node] = par[i - 1][par[i - 1][node]];
  for (int v : G[node])
    if (v != par[0][node]) {
      par[0][v] = node;
      h[v] = h[node] + 1;
      dfs1(v);
    }
  fnt[node] = timer;
}
inline bool is_par(int x, int y) {
  return stt[x] <= stt[y] && fnt[y] <= fnt[x];
}
int Lca(int x, int y) {
  if (h[x] > h[y]) swap(x, y);
  for (int i = 0; i < LOG; i++)
    if ((h[y] - h[x]) & (1 << i)) y = par[i][y];
  if (x == y) return x;
  for (int i = LOG - 1; ~i; i--)
    if (par[i][x] != par[i][y]) x = par[i][x], y = par[i][y];
  return par[0][x];
}
void Solve() {
  N = 0;
  cin >> k >> m;
  for (int i = 1; i <= k; i++) {
    cin >> V[i] >> S[i];
    A[N++] = V[i];
    vec[i].clear();
    vec[i].push_back(V[i]);
    limit[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    cin >> QQQ[i];
    A[N++] = QQQ[i];
  }
  sort(A, A + N, [](int i, int j) { return stt[i] < stt[j]; });
  N = unique(A, A + N) - A;
  int tmp = N;
  for (int i = 0; i + 1 < tmp; i++) A[N++] = Lca(A[i], A[i + 1]);
  sort(A, A + N, [](int i, int j) { return stt[i] < stt[j]; });
  N = unique(A, A + N) - A;
  for (int i = 0; i < N; i++)
    G2[A[i]].clear(), dist[A[i]] = dist2[A[i]] = inf,
                      mark[A[i]] = color[A[i]] = 0;
  stack<int> stk;
  for (int i = 0; i < N; i++) {
    int v = A[i];
    while (stk.size() && !is_par(stk.top(), v)) stk.pop();
    if (stk.size()) {
      int u = stk.top(), d = h[v] - h[u];
      G2[u].push_back({v, d});
      G2[v].push_back({u, d});
    }
    stk.push(v);
  }
  priority_queue<pair<pair<int, int>, int>, vector<pair<pair<int, int>, int>>,
                 greater<pair<pair<int, int>, int>>>
      pq;
  for (int i = 1; i <= k; i++)
    pq.push({{dist[V[i]] = dist2[V[i]] = 0, color[V[i]] = i}, V[i]});
  while (pq.size()) {
    int v = pq.top().second;
    pq.pop();
    if (mark[v]) continue;
    mark[v] = 1;
    for (pair<int, int> p : G2[v]) {
      int u = p.first, w = p.second,
          tmp = (dist2[v] + w + S[color[v]] - 1) / S[color[v]];
      if (pair<int, int>(tmp, color[v]) < pair<int, int>(dist[u], color[u])) {
        dist[u] = tmp;
        color[u] = color[v];
        dist2[u] = dist2[v] + w;
        pq.push({{dist[u], color[u]}, u});
      }
    }
  }
  for (int i = 1; i <= m; i++) cout << color[QQQ[i]] << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs1(1);
  cin >> q;
  while (q--) Solve();
  return 0;
}