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;
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
const int maxn = 2e5 + 10, maxm = 1e6 + 10, lg = 20, mod = 1e9 + 7, inf = 1e18;
int n, q, par[maxn][lg + 1], h[maxn], st[maxn], ft[maxn], tt, a[maxn], s[maxn],
    c[maxn];
pair<int, int> dis[maxn];
vector<int> g[maxn];
vector<pair<int, int>> gg[maxn];
bool mk[maxn];
bool cmp(int v, int u) { return st[v] > st[u]; }
void pfs(int v = 1, int p = 0) {
  st[v] = tt++;
  h[v] = h[p] + 1;
  par[v][0] = p;
  for (int i = 1; i <= lg; i++) par[v][i] = par[par[v][i - 1]][i - 1];
  for (auto u : g[v])
    if (u != p) pfs(u, v);
  ft[v] = tt;
}
bool ispar(int v, int u) { return st[v] <= st[u] && ft[v] >= ft[u]; }
int lca(int v, int u) {
  if (ispar(v, u)) return v;
  if (ispar(u, v)) return u;
  for (int i = lg; i >= 0; i--)
    if (par[v][i] != 0 && !ispar(par[v][i], u)) v = par[v][i];
  return par[v][0];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u), g[u].push_back(v);
  }
  pfs();
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> vv, sk;
    for (int i = 0; i < k; i++) cin >> a[i] >> s[i], vv.push_back(a[i]);
    for (int i = 0; i < m; i++) cin >> c[i], vv.push_back(c[i]);
    sort(vv.begin(), vv.end(), cmp),
        vv.resize(unique(vv.begin(), vv.end()) - vv.begin());
    for (int i = 0, tmp = (int)(vv.size()); i < tmp - 1; i++)
      vv.push_back(lca(vv[i], vv[i + 1]));
    sort(vv.begin(), vv.end(), cmp),
        vv.resize(unique(vv.begin(), vv.end()) - vv.begin());
    for (auto v : vv) mk[v] = 0, gg[v].clear(), dis[v] = {mod, mod};
    while (!vv.empty()) {
      while (!sk.empty() && !ispar(sk.back(), vv.back())) sk.pop_back();
      if (!sk.empty()) {
        int w = h[vv.back()] - h[sk.back()];
        gg[sk.back()].push_back({vv.back(), w});
        gg[vv.back()].push_back({sk.back(), w});
      }
      sk.push_back(vv.back());
      vv.pop_back();
    }
    priority_queue<pair<pair<int, int>, pair<int, int>>,
                   vector<pair<pair<int, int>, pair<int, int>>>,
                   greater<pair<pair<int, int>, pair<int, int>>>>
        pq;
    for (int i = 0; i < k; i++) pq.push({dis[a[i]] = {0, i}, {0, a[i]}});
    while (!pq.empty()) {
      int v = pq.top().second.second, i = pq.top().first.second,
          d = pq.top().second.first;
      pq.pop();
      if (mk[v]) continue;
      mk[v] = 1;
      for (auto uw : gg[v]) {
        int u = uw.first, w = uw.second;
        pair<int, int> xx = {(d + w + s[i] - 1) / s[i], i};
        if (dis[u] > xx) pq.push({dis[u] = xx, {d + w, u}});
      }
    }
    for (int i = 0; i < m; i++) cout << dis[c[i]].second + 1 << ' ';
    cout << '\n';
  }
  return 0;
}