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;
const int maxn = 2e5, maxm = 4e5;
int n, Q, k, m, cur, dfn[maxn + 3], dep[maxn + 3], fa[maxn + 3], id[maxn + 3],
    sz[maxn + 3], son[maxn + 3];
int city[maxm + 3], speed[maxn + 3], imp[maxm + 3], st[maxn + 3], top,
    infe[maxn + 3], vis[maxn + 3];
vector<int> G[maxn + 3], T[maxn + 3];
void dfs1(int u, int f = 0) {
  dfn[u] = ++cur, sz[u] = 1;
  for (int i = 0, v; i < G[u].size(); i++) {
    v = G[u][i];
    if (v == f) continue;
    dep[v] = dep[u] + 1, dfs1(v, u);
    fa[v] = u, sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int f = 0, int i = 1) {
  id[u] = i;
  if (son[u]) dfs2(son[u], u, i);
  for (int i = 0, v; i < G[u].size(); i++) {
    v = G[u][i];
    if (v == f || v == son[u]) continue;
    dfs2(v, u, v);
  }
}
int lca(int u, int v) {
  while (id[u] != id[v]) {
    if (dep[id[u]] < dep[id[v]]) swap(u, v);
    u = fa[id[u]];
  }
  return dep[u] < dep[v] ? u : v;
}
bool comp(int i, int j) { return dfn[i] < dfn[j]; }
void clear(int u) {
  if (vis[u] != cur) {
    vis[u] = cur, vector<int>().swap(T[u]), infe[u] = 0;
  }
}
void add(int u, int v) {
  clear(u), clear(v), T[u].push_back(v), T[v].push_back(u);
}
int getdis(int u, int v) {
  return dep[u] - dep[v] < 0 ? dep[v] - dep[u] : dep[u] - dep[v];
}
void insert(int u) {
  int x = lca(u, st[top]);
  while (top >= 2 && dep[st[top - 1]] >= dep[x])
    add(st[top], st[top - 1]), top--;
  if (st[top] != x) add(st[top], x), st[top] = x;
  if (st[top] != u) st[++top] = u;
}
struct node {
  int x, ti, id, dis;
  node(int _x = 0, int t = 0, int i = 0, int d = 0) {
    x = _x, ti = t, id = i, dis = d;
  }
  friend bool operator<(const node &a, const node &b) {
    return a.ti == b.ti ? a.id > b.id : a.ti > b.ti;
  }
};
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    G[x].push_back(y), G[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(1, 1);
  scanf("%d", &Q);
  cur = 0;
  while (Q--) {
    scanf("%d %d", &k, &m), cur++;
    for (int i = 1; i <= k; i++)
      scanf("%d %d", &city[i], &speed[i]), imp[i] = city[i];
    for (int i = k + 1; i <= k + m; i++)
      scanf("%d", &city[i]), imp[i] = city[i];
    sort(imp + 1, imp + 1 + k + m,
         [&](const int a, const int b) { return dfn[a] < dfn[b]; });
    top = 0, st[++top] = 1, imp[0] = 1, vector<int>().swap(T[1]), infe[1] = 0;
    for (int i = 1; i <= k + m; i++)
      if (imp[i] != imp[i - 1]) insert(imp[i]);
    for (int i = top; i >= 2; i--) add(st[i], st[i - 1]);
    priority_queue<node> q;
    for (int i = 1; i <= k; i++) q.push((node){city[i], 0, i, 0});
    while (q.size()) {
      node t = q.top();
      q.pop();
      if (infe[t.x]) continue;
      infe[t.x] = t.id;
      for (int y : T[t.x]) {
        int totdis = t.dis + getdis(t.x, y);
        if (!infe[y])
          q.push((node){y, (int)ceil(1. * totdis / speed[t.id]), t.id, totdis});
      }
    }
    for (int i = k + 1; i <= k + m; i++)
      printf("%d%c", infe[city[i]], " \n"[i == k + m]);
  }
  return 0;
}