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 = 2e5 + 500;
const int M = 1e6 + 500;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int LOG = 18;
vector<int> v[N], v2[N];
vector<int> skup, poc, pitam;
int par[N][LOG], dep[N], ans[N], n;
int tme[N], R[N], cur_tme, sped[N];
bool cmp(int x, int y) { return tme[x] < tme[y]; }
void dfs(int x, int lst) {
  tme[x] = cur_tme++;
  par[x][0] = lst;
  dep[x] = dep[lst] + 1;
  for (int y : v[x])
    if (y != lst) dfs(y, x);
  R[x] = cur_tme - 1;
}
void precompute() {
  for (int i = 1; i < LOG; i++) {
    for (int j = 1; j <= n; j++) par[j][i] = par[par[j][i - 1]][i - 1];
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = LOG - 1; i >= 0; i--)
    if (dep[x] - (1 << i) >= dep[y]) x = par[x][i];
  if (x == y) return x;
  for (int i = LOG - 1; i >= 0; i--)
    if (par[x][i] != par[y][i]) x = par[x][i], y = par[y][i];
  return par[x][0];
}
inline int dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; }
void prosiri() {
  sort(skup.begin(), skup.end(), cmp);
  int ln = (int)skup.size();
  for (int i = 1; i < ln; i++) {
    skup.push_back(lca(skup[i], skup[i - 1]));
  }
  sort(skup.begin(), skup.end());
  skup.erase(unique(skup.begin(), skup.end()), skup.end());
}
void pospajaj() {
  stack<int> tren;
  sort(skup.begin(), skup.end(), cmp);
  for (int i = 0; i < (int)skup.size(); i++) {
    while (!tren.empty() && R[tren.top()] < tme[skup[i]]) tren.pop();
    if (!tren.empty()) {
      v2[skup[i]].push_back(tren.top());
      v2[tren.top()].push_back(skup[i]);
    }
    tren.push(skup[i]);
  }
}
void obradi() {
  set<pair<pair<int, pair<int, int> >, int> > S;
  for (int i = 0; i < (int)poc.size(); i++) {
    S.insert({{0, {i + 1, 0}}, poc[i]});
  }
  for (; !S.empty();) {
    pair<pair<int, pair<int, int> >, int> sad = *S.begin();
    S.erase(S.begin());
    int cur = sad.second;
    int ost = sad.first.second.second, boj = sad.first.second.first;
    if (ans[cur]) continue;
    ans[cur] = boj;
    for (int sus : v2[cur]) {
      if (ans[sus]) continue;
      int ds = dis(cur, sus);
      S.insert(
          {{(ost + ds + sped[boj] - 1) / sped[boj], {boj, (ost + ds)}}, sus});
    }
  }
}
void pocisti() {
  for (int x : skup) {
    v2[x].clear();
    ans[x] = 0;
  }
  poc.clear();
  pitam.clear();
  skup.clear();
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y), v[y].push_back(x);
  }
  dfs(1, 1);
  precompute();
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int k, m;
    scanf("%d%d", &k, &m);
    for (int j = 1; j <= k; j++) {
      int x;
      scanf("%d%d", &x, &sped[j]);
      poc.push_back(x);
      skup.push_back(x);
    }
    for (int j = 0; j < m; j++) {
      int x;
      scanf("%d", &x);
      pitam.push_back(x);
      skup.push_back(x);
    }
    prosiri();
    pospajaj();
    obradi();
    for (int x : pitam) printf("%d ", ans[x]);
    printf("\n");
    pocisti();
  }
  return 0;
}