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
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void cmin(T &a, T b) {
  ((a > b) && (a = b));
}
template <class T>
inline void cmax(T &a, T b) {
  ((a < b) && (a = b));
}
char IO;
template <class T = int>
T rd() {
  T s = 0;
  int f = 0;
  while (!isdigit(IO = getchar())) f |= IO == '-';
  do s = (s << 1) + (s << 3) + (IO ^ '0');
  while (isdigit(IO = getchar()));
  return f ? -s : s;
}
const int N = 4e5 + 10, INF = 1e9 + 10;
int n, m, k;
vector<int> G[N], E[N];
int L[N], R[N], I[N], dfn, fa[N][20], dep[N];
void dfs(int u, int f) {
  I[L[u] = ++dfn] = u;
  dep[u] = dep[fa[u][0] = f] + 1;
  for (int i = 1, iend = 19; i <= iend; ++i) fa[u][i] = fa[fa[u][i - 1]][i - 1];
  for (int v : G[u])
    if (v != f) dfs(v, u);
  R[u] = dfn;
}
int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 0, del = dep[x] - dep[y]; (1 << i) <= del; ++i)
    if (del & (1 << i)) x = fa[x][i];
  if (x == y) return x;
  for (int i = 18, iend = 0; i >= iend; --i)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
int vis[N], S[N], T;
void Link(int u, int v) { E[u].push_back(v), E[v].push_back(u); }
void Ins(int u) {
  if (S[T] == u) return;
  E[u].clear(), vis[u] = 0;
  if (!T) return void(S[++T] = u);
  int lca = LCA(u, S[T]);
  if (lca == S[T]) return void(S[++T] = u);
  while (T > 1 && L[S[T - 1]] >= L[lca]) Link(S[T], S[T - 1]), T--;
  if (S[T] != lca) vis[lca] = 0, E[lca].clear(), Link(lca, S[T]), S[T] = lca;
  S[++T] = u;
}
int P[N], B[N], Q[N], A[N * 2], C;
struct Node {
  int u, f, d;
  long long ti() const { return 1ll * (d + B[f] - 1) / B[f] * (k + 1) + f; }
  bool operator<(const Node __) const { return ti() > __.ti(); }
};
priority_queue<Node> que;
void Solve() {
  k = rd(), m = rd(), C = 0;
  for (int i = 1, iend = k; i <= iend; ++i) A[++C] = P[i] = rd(), B[i] = rd();
  for (int i = 1, iend = m; i <= iend; ++i) A[++C] = Q[i] = rd();
  sort(A + 1, A + C + 1, [&](int x, int y) { return L[x] < L[y]; }), T = 0;
  for (int i = 1, iend = C; i <= iend; ++i) Ins(A[i]);
  while (T > 1) Link(S[T - 1], S[T]), T--;
  for (int i = 1, iend = k; i <= iend; ++i) que.push((Node){P[i], i, 0});
  while (!que.empty()) {
    Node u = que.top();
    que.pop();
    if (vis[u.u]) continue;
    vis[u.u] = u.f;
    for (int v : E[u.u])
      if (!vis[v]) que.push((Node){v, u.f, u.d + abs(dep[u.u] - dep[v])});
  }
  for (int i = 1, iend = m; i <= iend; ++i) printf("%d ", vis[Q[i]]);
  puts("");
}
int main() {
  n = rd();
  for (int i = 2, iend = n; i <= iend; ++i) {
    int u = rd(), v = rd();
    G[u].push_back(v), G[v].push_back(u);
  }
  dfs(1, 0);
  for (int _ = 1, _end = rd(); _ <= _end; ++_) Solve();
}
