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 ll = long long;
using ld = long double;
using namespace std;
const ll N = 400001, L = 19;
vector<ll> g[N];
vector<pair<ll, ll>> ng[N];
ll tin[N], tout[N], ct, ord[N], fv[N], s[N], qq[N], h[N], up[L][N], vir[N],
    ft[N], k;
set<vector<ll>> qs;
void dfs(ll v, ll p) {
  ord[ct] = v;
  tin[v] = ct++;
  for (ll u : g[v]) {
    if (u == p) continue;
    h[u] = h[v] + 1;
    up[0][u] = v;
    dfs(u, v);
  }
  tout[v] = ct;
}
ll lca(ll u, ll v) {
  if (h[u] < h[v]) swap(u, v);
  for (ll i = L - 1; i >= 0; i--)
    if (h[up[i][u]] >= h[v]) u = up[i][u];
  assert(h[u] == h[v]);
  if (u == v) return u;
  for (ll i = L - 1; i >= 0; i--)
    if (up[i][u] != up[i][v]) {
      u = up[i][u];
      v = up[i][v];
    }
  return up[0][u];
}
void add(ll v, ll ml) {
  for (auto &pp : ng[v]) {
    if (vir[pp.first] != -1) continue;
    if (ml >= pp.second)
      qs.insert({ft[v], pp.first, ml - pp.second});
    else {
      ll dt = (pp.second - ml + s[vir[v]] - 1) / s[vir[v]];
      qs.insert({ft[v] + dt * k, pp.first, dt * s[vir[v]] + ml - pp.second});
    }
  }
}
ll isp(ll u, ll v) { return tin[u] <= tin[v] && tout[v] <= tout[u]; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ll n, u, v, q, m;
  cin >> n;
  for (ll i = 0; i < (n - 1); i++) {
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, 0);
  for (ll i = 0; i < (L - 1); i++)
    for (ll j = 0; j < (n); j++) up[i + 1][j] = up[i][up[i][j]];
  cin >> q;
  for (ll i = 0; i < (n); i++) vir[i] = -1;
  for (ll z = 0; z < (q); z++) {
    cin >> k >> m;
    vector<ll> li;
    for (ll i = 0; i < (k); i++) {
      cin >> fv[i] >> s[i];
      fv[i]--;
      vir[fv[i]] = i;
      ft[fv[i]] = i;
      li.push_back(tin[fv[i]]);
    }
    for (ll i = 0; i < (m); i++) {
      cin >> qq[i];
      qq[i]--;
      li.push_back(tin[qq[i]]);
    }
    sort((li).begin(), (li).end());
    li.resize(unique((li).begin(), (li).end()) - li.begin());
    ll cs = li.size();
    for (ll i = 0; i < (cs); i++) {
      v = ord[li[i]];
      ll l1 = lca(v, ord[li[(i + 1) % cs]]);
      li.push_back(tin[l1]);
    }
    sort((li).begin(), (li).end());
    li.resize(unique((li).begin(), (li).end()) - li.begin());
    vector<ll> st;
    int cnt = 0;
    for (ll i : li) {
      while (!st.empty() && !isp(ord[st.back()], ord[i])) st.pop_back();
      if (!st.empty()) {
        v = ord[i];
        u = ord[st.back()];
        ng[v].push_back({u, h[v] - h[u]});
        ng[u].push_back({v, h[v] - h[u]});
        cnt++;
      }
      st.push_back(i);
    }
    assert(cnt + 1 == li.size());
    for (ll i = 0; i < (k); i++) add(fv[i], 0);
    while (!qs.empty()) {
      ll ct = (*qs.begin())[0], cv = (*qs.begin())[1], cl = (*qs.begin())[2];
      qs.erase(qs.begin());
      if (vir[cv] != -1) continue;
      vir[cv] = ct % k;
      ft[cv] = ct;
      add(cv, cl);
    }
    for (ll i = 0; i < (m); i++) cout << vir[qq[i]] + 1 << ' ';
    cout << '\n';
    for (ll i : li) {
      v = ord[i];
      ng[v].clear();
      vir[v] = -1;
      ft[v] = 0;
    }
  }
  return 0;
}