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;
template <typename T>
inline T read(register T& t) {
  register T f = 1;
  register char ch = getchar();
  t = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') t = t * 10 + ch - '0', ch = getchar();
  t *= f;
  return t;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
const long long p = 998244353;
inline long long power(register long long x, register long long k = p - 2) {
  register long long re = 1;
  for (; k; k >>= 1, x = x * x % p)
    if (k & 1) re = re * x % p;
  return re;
}
int n, Q;
namespace T1 {
int head[500005], to[1000005], ne[10000005], total = 1;
inline void add(int a, int b) {
  total++;
  to[total] = b;
  ne[total] = head[a];
  head[a] = total;
}
int deep[500005], size[500005];
int son[500005], fa[500005];
int top[500005], id[500005], id_top;
void dfs1(int now) {
  deep[now] = deep[fa[now]] + 1;
  size[now] = 1;
  for (int i = head[now]; i; i = ne[i])
    if (to[i] != fa[now]) {
      fa[to[i]] = now;
      dfs1(to[i]);
      size[now] += size[to[i]];
      if (size[to[i]] > size[son[now]]) son[now] = to[i];
    }
}
void dfs2(int now, int topf) {
  id[now] = ++id_top;
  top[now] = topf;
  if (son[now]) dfs2(son[now], topf);
  for (int i = head[now]; i; i = ne[i])
    if (to[i] != fa[now] && to[i] != son[now]) dfs2(to[i], to[i]);
}
inline int lca(int x, int y) {
  while (top[x] != top[y]) {
    if (deep[top[x]] < deep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  if (deep[x] < deep[y]) return x;
  return y;
}
inline int dis(int x, int y) { return deep[x] + deep[y] - deep[lca(x, y)] * 2; }
}  // namespace T1
int Source[500005], Speed[500005];
int Query[500005];
namespace T2 {
int head[500005], ne[1000005], to[1000005], total = 1;
inline void add(int a, int b) {
  total++;
  to[total] = b;
  ne[total] = head[a];
  head[a] = total;
}
}  // namespace T2
int main() {
  read(n);
  for (int i = 1, x, y; i < n; i++) read(x, y), T1::add(x, y), T1::add(y, x);
  T1::dfs1(1);
  T1::dfs2(1, 1);
  read(Q);
  while (Q--) {
    static int pot[1000005];
    int pot_top = 0;
    int K, M;
    read(K, M);
    for (int i = 1; i <= K; i++)
      read(Source[i], Speed[i]), pot[++pot_top] = Source[i];
    for (int i = 1; i <= M; i++) read(Query[i]), pot[++pot_top] = Query[i];
    sort(pot + 1, pot + pot_top + 1);
    pot_top = unique(pot + 1, pot + pot_top + 1) - pot - 1;
    sort(pot + 1, pot + pot_top + 1,
         [&](int a, int b) { return T1::id[a] < T1::id[b]; });
    static int st[500005];
    int st_top = 0;
    static int rev[500005];
    int rev_top = 0;
    st[st_top = 1] = 1;
    rev[rev_top = 1] = 1;
    for (int i = pot[1] == 1 ? 2 : 1; i <= pot_top; i++) {
      int x = pot[i], y = T1::lca(st[st_top], x);
      while (st_top > 1 && T1::id[y] <= T1::id[st[st_top - 1]])
        T2::add(st[st_top], st[st_top - 1]),
            T2::add(st[st_top - 1], st[st_top]), rev[++rev_top] = st[st_top--];
      if (st[st_top] != y)
        T2::add(st[st_top], y), T2::add(y, st[st_top]),
            rev[++rev_top] = st[st_top], st[st_top] = y;
      st[++st_top] = x;
    }
    while (st_top > 1)
      T2::add(st[st_top], st[st_top - 1]), T2::add(st[st_top - 1], st[st_top]),
          rev[++rev_top] = st[st_top--];
    static tuple<int, int, int, int> dis[500005];
    dis[1] = make_tuple(-1e9, -1e9, -1e9, -1e9);
    static int Ans[500005];
    Ans[1] = 0;
    for (int i = 1; i <= rev_top; i++)
      dis[rev[i]] = make_tuple(-1e9, -1e9, -1e9, -1e9), Ans[rev[i]] = 0;
    priority_queue<tuple<int, int, int, int> > q;
    for (int i = 1; i <= K; i++)
      dis[Source[i]] = max(dis[Source[i]], make_tuple(0, -i, 0, Source[i])),
      q.push(make_tuple(0, -i, 0, Source[i]));
    while (q.size()) {
      int now = get<3>(q.top());
      int from = -get<1>(q.top());
      q.pop();
      if (Ans[now]) continue;
      Ans[now] = from;
      for (int i = T2::head[now]; i; i = T2::ne[i]) {
        int v = T2::to[i], Len = T1::dis(Source[from], v), len;
        if (Len % Speed[from])
          len = Len / Speed[from] + 1;
        else
          len = Len / Speed[from];
        int les = Speed[from] * len - Len;
        tuple<int, int, int, int> qwq = make_tuple(-len, -from, les, v);
        if (qwq > dis[v]) q.push(dis[v] = qwq);
      }
    }
    for (int i = 1; i <= M; i++) printf("%d ", Ans[Query[i]]);
    printf("\n");
    T2::total = 1;
    for (int i = 1; i <= rev_top; i++) T2::head[rev[i]] = 0;
  }
  return 0;
}