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 N = 4e5 + 50, inf = 0x3f3f3f3f;
int n, q, f[N], sz[N], tp[N], d[N], dfn[N], clk, k, m, vir[N], spd[N], cit[N],
    c[N], tt, st[N], tpp, ss[N], pp;
int id[N], tim[N], rem[N];
vector<int> e[N];
int rd() {
  int x = 0, c;
  while (!isdigit(c = getchar()))
    ;
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return x;
}
void dfs1(int x) {
  if (f[x]) e[x].erase(find(e[x].begin(), e[x].end(), f[x]));
  sz[x] = 1;
  for (int i = 0, y; i < e[x].size(); i++) {
    f[y = e[x][i]] = x;
    d[y] = d[x] + 1;
    dfs1(y), sz[x] += sz[y];
    sz[y] > sz[e[x][0]] ? swap(e[x][i], e[x][0]), 0 : 0;
  }
}
void dfs2(int x) {
  dfn[x] = ++clk;
  for (int i = 0; i < e[x].size(); i++)
    tp[e[x][i]] = i ? e[x][i] : tp[x], dfs2(e[x][i]);
}
int lca(int x, int y) {
  for (; tp[x] != tp[y]; x = f[tp[x]])
    if (d[tp[x]] < d[tp[y]]) swap(x, y);
  return d[x] < d[y] ? x : y;
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
void extend(int x) {
  int y = lca(x, st[tpp]);
  while (tpp > 1 && d[st[tpp - 1]] >= d[y]) {
    e[st[tpp]].push_back(st[tpp - 1]);
    e[st[tpp - 1]].push_back(st[tpp]);
    tpp--;
  }
  if (st[tpp] != y)
    e[y].push_back(st[tpp]), e[st[tpp]].push_back(y), st[tpp] = y;
  st[++tpp] = x;
}
void update(int x, int y) {
  if (tim[y] > tim[x] || tim[y] == inf) return;
  int dis = abs(d[y] - d[x]),
      t = tim[y] +
          (rem[y] < dis ? (dis - rem[y] + spd[id[y]] - 1) / spd[id[y]] : 0);
  if (t < tim[x] || id[y] < id[x] && t == tim[x]) {
    tim[x] = t;
    id[x] = id[y];
    rem[x] =
        t > tim[y] ? (t - tim[y]) * spd[id[y]] - (dis - rem[y]) : rem[y] - dis;
  }
}
void up(int x, int ff) {
  ss[++pp] = x;
  if (ff) e[x].erase(find(e[x].begin(), e[x].end(), ff));
  for (int i = 0, y; i < e[x].size(); i++) up(y = e[x][i], x), update(x, y);
}
void down(int x) {
  for (int i = 0, y; i < e[x].size(); i++) update(y = e[x][i], x), down(y);
}
int main() {
  n = rd();
  for (int i = 1, x, y; i < n; i++)
    x = rd(), y = rd(), e[x].push_back(y), e[y].push_back(x);
  dfs1(1);
  tp[1] = 1;
  dfs2(1);
  q = rd();
  for (int i = 1; i <= n; i++) e[i].clear();
  memset(tim, 0x3f, sizeof(tim));
  for (int i = 1; i <= q; i++) {
    k = rd();
    m = rd();
    c[tt = 1] = 1;
    for (int j = 1; j <= k; j++) c[++tt] = vir[j] = rd(), spd[j] = rd();
    for (int j = 1; j <= m; j++) c[++tt] = cit[j] = rd();
    sort(c + 1, c + tt + 1, cmp);
    tt = unique(c + 1, c + tt + 1) - c - 1;
    st[tpp = 1] = 1;
    for (int j = 2; j <= tt; j++) extend(c[j]);
    while (tpp > 1)
      e[st[tpp]].push_back(st[tpp - 1]), e[st[tpp - 1]].push_back(st[tpp]),
          tpp--;
    for (int j = 1; j <= k; j++)
      tim[vir[j]] = 0, id[vir[j]] = j, rem[vir[j]] = 0;
    up(1, 0);
    down(1);
    for (int j = 1; j <= m; j++) printf("%d%c", id[cit[j]], " \n"[j == m]);
    while (pp) e[ss[pp]].clear(), tim[ss[pp--]] = inf;
  }
  return 0;
}