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 gi() {
  T x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const int N = 200003, M = N << 3;
int n, Q;
int tot, head[N], ver[M], nxt[M];
int tp, stk[M];
int tmp[M], cnt;
int qry[N];
int fa[N], son[N], sz[N], topp[N], dep[N], dfn[N], tim;
struct Node {
  int tim, id, s, sy, num;
  bool operator<(const Node &x) const {
    if (tim != x.tim) return tim > x.tim;
    if (id != x.id) return id > x.id;
    return sy < x.sy;
  }
} a[N];
priority_queue<Node> q;
vector<int> g[N];
inline void add(int u, int v) {
  ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}
inline void add1(int u, int v) { g[u].push_back(v), g[v].push_back(u); }
namespace LCA {
void dfs1(int u, int f) {
  fa[u] = f, sz[u] = 1, dep[u] = dep[f] + 1, dfn[u] = ++tim;
  for (int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == f) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int f) {
  topp[u] = f;
  if (!son[u]) return;
  dfs2(son[u], f);
  for (int i = head[u]; i; i = nxt[i]) {
    int v = ver[i];
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v, v);
  }
}
int getLCA(int u, int v) {
  while (topp[u] != topp[v]) {
    if (dep[topp[u]] < dep[topp[v]]) swap(u, v);
    u = fa[topp[u]];
  }
  return dep[u] < dep[v] ? u : v;
}
}  // namespace LCA
inline int dis(int u, int v) {
  return dep[u] + dep[v] - (dep[LCA ::getLCA(u, v)] << 1);
}
inline bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
inline void Ins(int v) {
  if (tp <= 1) {
    stk[++tp] = v;
    return;
  }
  int u = stk[tp];
  int lca = LCA ::getLCA(u, v);
  if (lca == u) {
    stk[++tp] = v;
    return;
  }
  while (tp > 1 && dfn[stk[tp - 1]] > dfn[lca])
    add1(stk[tp - 1], stk[tp]), --tp;
  if (lca == stk[tp]) {
    if (tp > 1) add1(stk[tp - 1], stk[tp]);
  } else
    add1(lca, stk[tp]), stk[tp] = lca;
  stk[++tp] = v;
  return;
}
bool vis[N];
inline Node Goto(int st, int ed) {
  int d = dis(st, ed);
  Node tmp = a[st];
  if (tmp.sy > d) {
    tmp.sy -= d;
    return tmp;
  }
  d -= a[st].sy;
  int deltat = d / a[st].s + (d % a[st].s != 0);
  tmp.tim += deltat, tmp.sy = deltat * a[st].s - d;
  return tmp;
}
inline void Dij() {
  while (!q.empty()) {
    auto x = q.top();
    q.pop();
    if (vis[x.num]) continue;
    vis[x.num] = true;
    for (auto v : g[x.num]) {
      Node tmp = Goto(x.num, v);
      if (a[v] < tmp) {
        a[v] = tmp;
        a[v].num = v;
        q.push(a[v]);
      }
    }
  }
}
void Clear(int u, int f) {
  for (auto x : g[u])
    if (x != u && x != f) Clear(x, u);
  g[u].clear(), a[u] = {n + 1, n + 1, 0, 0, u}, vis[u] = false;
}
int main() {
  n = gi<int>();
  for (int i = 1; i < n; i += 1) {
    int u = gi<int>(), v = gi<int>();
    add(u, v), add(v, u);
  }
  LCA ::dfs1(1, 0);
  LCA ::dfs2(1, 1);
  for (int i = 1; i <= n; i += 1) a[i] = {n + 1, n + 1, 0, 0, i};
  Q = gi<int>();
  while (Q--) {
    int ki = gi<int>(), mi = gi<int>();
    cnt = ki;
    for (int i = 1; i <= ki; i += 1) {
      tmp[i] = gi<int>(), a[tmp[i]].id = i, a[tmp[i]].s = gi<int>(),
      a[tmp[i]].tim = 0, a[tmp[i]].sy = 0, a[tmp[i]].num = tmp[i];
      q.push(a[tmp[i]]);
    }
    for (int i = 1; i <= mi; i += 1) qry[i] = tmp[++cnt] = gi<int>();
    sort(tmp + 1, tmp + 1 + cnt, cmp);
    tp = 0;
    stk[tp = 1] = tmp[0] = 1;
    for (int i = 1; i <= cnt; i += 1)
      if (tmp[i] != tmp[i - 1]) Ins(tmp[i]);
    for (int i = 1; i < tp; i += 1)
      if (stk[i] != stk[i + 1]) add1(stk[i], stk[i + 1]);
    Dij();
    for (int i = 1; i <= mi; i += 1) printf("%d ", a[qry[i]].id);
    puts("");
    Clear(1, 0);
  }
  return 0;
}