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;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 400010;
const int INF = 1e9;
int N;
pair<int, int> T[MAXN + 1];
int Node[MAXN << 1], Next[MAXN << 1], Root[MAXN + 1], cnt;
int top, sta[MAXN + 1], In[MAXN + 1], all;
int Spd[MAXN + 1], col[MAXN + 1];
int dep[MAXN + 1];
vector<int> vec[MAXN + 1];
int pre[20][MAXN + 1];
inline int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  for (int i = 19; i >= 0; i--)
    if (dep[u] - dep[v] >= (1 << i)) u = pre[i][u];
  for (int i = 19; i >= 0; i--)
    if (pre[i][u] != pre[i][v]) u = pre[i][u], v = pre[i][v];
  if (u != v) return pre[0][u];
  return u;
}
int fa[MAXN + 1], dfn[MAXN + 1];
inline void link(int x, int y) {
  vec[x].push_back(y);
  fa[y] = x;
}
inline void extend(int x) {
  if (x == sta[top]) return;
  if (!top) {
    sta[++top] = x;
    In[++all] = x;
    return;
  }
  int y = LCA(sta[top], x);
  while (top > 1 && dep[sta[top - 1]] >= dep[y])
    link(sta[top - 1], sta[top]), --top;
  if (sta[top] != y) link(y, sta[top]);
  sta[top] = y;
  sta[++top] = x;
  In[++all] = y;
  In[++all] = x;
}
inline void insert(int u, int v) {
  Node[++cnt] = v;
  Next[cnt] = Root[u];
  Root[u] = cnt;
}
struct data {
  int k, s;
} q[MAXN + 1];
inline bool cmp(data a, data b) { return dfn[a.k] < dfn[b.k]; }
int tim;
inline void dfs(int k, int Fa) {
  pre[0][k] = Fa;
  for (int j = 1; j <= 19; j++) pre[j][k] = pre[j - 1][pre[j - 1][k]];
  dep[k] = dep[Fa] + 1;
  dfn[k] = ++tim;
  for (int x = Root[k]; x; x = Next[x]) {
    int v = Node[x];
    if (v == Fa) continue;
    dfs(v, k);
  }
}
pair<int, int> upd(pair<int, int> a, pair<int, int> b) {
  if (!a.second) return b;
  if (!b.second) return a;
  int ta = ((a.first + Spd[a.second] - 1) / Spd[a.second]),
      tb = ((b.first + Spd[b.second] - 1) / Spd[b.second]);
  if (ta == tb) {
    if (a.second == b.second) {
      if (a.first < b.first) return a;
      return b;
    }
    if (col[a.second] < col[b.second]) return a;
    return b;
  }
  if (ta < tb)
    return a;
  else
    return b;
}
inline void dfs1(int k) {
  T[k] = make_pair(INF, 0);
  if (col[k]) T[k] = make_pair(0, k);
  for (int i = 0, sz = vec[k].size(); i < sz; i++) {
    int v = vec[k][i];
    dfs1(v);
    pair<int, int> tson = T[v];
    tson.first += dep[v] - dep[k];
    T[k] = upd(T[k], tson);
  }
}
inline void dfs2(int k) {
  pair<int, int> tfa = T[fa[k]];
  tfa.first += dep[k] - dep[fa[k]];
  if (fa[k]) T[k] = upd(T[k], tfa);
  for (int i = 0, sz = vec[k].size(); i < sz; i++) {
    int v = vec[k][i];
    dfs2(v);
  }
}
int Ans[MAXN + 1];
int main() {
  N = read();
  Spd[0] = 1;
  for (int i = 1; i < N; i++) {
    int u = read(), v = read();
    insert(u, v);
    insert(v, u);
  }
  dfs(1, 0);
  int Q = read();
  while (Q--) {
    int K = read(), M = read();
    top = 0;
    bool flag = 0;
    for (int i = 1; i <= K; i++) {
      q[i].k = read(), q[i].s = read();
      flag |= (q[i].k == 1);
      col[q[i].k] = i;
      Spd[q[i].k] = q[i].s;
    }
    for (int i = 1; i <= M; i++) {
      q[++K].k = read(), q[K].s = -i, flag |= (q[K].k == 1);
      if (col[q[K].k]) {
        Ans[i] = col[q[K].k];
        --K;
      }
    }
    if (!flag) q[++K].k = 1, q[K].s = 0;
    sort(q + 1, q + K + 1, cmp);
    for (int i = 1; i <= K; i++) extend(q[i].k);
    fa[1] = 0;
    while (top > 1) link(sta[top - 1], sta[top]), --top;
    top = 0;
    dfs1(1);
    dfs2(1);
    for (int i = 1; i <= K; i++) {
      if (q[i].s < 0) Ans[-q[i].s] = col[T[q[i].k].second];
    }
    for (int i = 1; i <= M; i++) printf("%d ", Ans[i]);
    printf("\n");
    for (int i = 1; i <= all; i++)
      vec[In[i]].clear(), col[In[i]] = Spd[In[i]] = 0, fa[i] = 0;
    all = 0;
  }
  return 0;
}