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 sz = 2e5 + 10;
vector<int> sv[sz];
int tin[sz], timer = 0, up[sz][20], h[sz];
void dfs(int v) {
  tin[v] = timer++;
  for (int a = 0; a < sv[v].size(); a++) {
    int ne = sv[v][a];
    if (ne != up[v][0]) up[ne][0] = v, h[ne] = h[v] + 1, dfs(ne);
  }
}
int lca(int u, int v) {
  if (tin[u] > tin[v]) swap(u, v);
  for (int i = 19; i >= 0; i--) {
    int ne = up[v][i];
    if (tin[ne] > tin[u]) v = ne;
  }
  if (v != u) v = up[v][0];
  return v;
}
int s[sz];
pair<int, int> di[sz];
vector<pair<int, int> > tr[sz];
bool comp(int a, int b) { return tin[a] < tin[b]; }
bool comp2(pair<int, int> a, pair<int, int> b) {
  int f1 = a.first / s[a.second], f2 = b.first / s[b.second];
  if (f1 != f2) return f1 < f2;
  if (a.second != b.second) return a.second < b.second;
  return a.first < b.first;
}
bool comp3(int a, int b) {
  if (comp2(di[a], di[b])) return true;
  if (comp2(di[b], di[a])) return false;
  return a < b;
}
int main() {
  s[sz - 1] = 1;
  int n;
  cin >> n;
  for (int a = 0; a < n - 1; a++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--, v--;
    sv[u].push_back(v);
    sv[v].push_back(u);
  }
  dfs(0);
  for (int i = 1; i < 20; i++) {
    for (int a = 0; a < n; a++) up[a][i] = up[up[a][i - 1]][i - 1];
  }
  int q;
  cin >> q;
  for (int a = 0; a < q; a++) {
    int k, m;
    scanf("%d%d", &k, &m);
    int vi[k], que[m];
    vector<int> sp, st(1, 0), ve(1, 0);
    for (int b = 0; b < k; b++) {
      scanf("%d%d", &vi[b], &s[b]);
      vi[b]--, sp.push_back(vi[b]);
    }
    for (int b = 0; b < m; b++) {
      scanf("%d", &que[b]);
      que[b]--, sp.push_back(que[b]);
    }
    sort(sp.begin(), sp.end(), comp);
    sp.push_back(0);
    for (int b = 0; b < sp.size(); b++) {
      int u = sp[b], lc = lca(u, st.back());
      while (h[st.back()] > h[lc]) {
        int v = st.back();
        ve.push_back(v);
        st.pop_back();
        int pr = st.back();
        if (h[pr] < h[lc]) pr = lc;
        tr[v].push_back({pr, h[v] - h[pr]});
        tr[pr].push_back({v, h[v] - h[pr]});
      }
      if (st.back() != lc) st.push_back(lc);
      if (st.back() != u) st.push_back(u);
    }
    for (int b = 0; b < ve.size(); b++) di[ve[b]] = {1e9, sz - 1};
    multiset<int, bool (*)(int, int)> se(&comp3);
    for (int b = 0; b < k; b++) {
      di[vi[b]] = {s[b] - 1, b};
      se.insert(vi[b]);
    }
    while (se.size()) {
      int v = *se.begin();
      se.erase(se.begin());
      for (int a = 0; a < tr[v].size(); a++) {
        int ne = tr[v][a].first;
        pair<int, int> va = di[v];
        va.first += tr[v][a].second;
        if (comp2(va, di[ne])) {
          se.erase(ne);
          di[ne] = va;
          se.insert(ne);
        }
      }
    }
    for (int b = 0; b < m; b++) printf("%d ", di[que[b]].second + 1);
    printf("\n");
    for (int b = 0; b < ve.size(); b++) tr[ve[b]].clear();
  }
}