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;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
using ll = long long;
using ld = long double;
const string FILENAME = "input";
const int MAXN = 400228;
int n;
vector<int> g[MAXN];
int vs[MAXN], s[MAXN];
vector<int> st;
int z[MAXN];
int in[MAXN], out[MAXN];
int timer = 0;
int deep[MAXN];
int jump[22][MAXN];
void dfs(int u, int pr = -1) {
  timer++;
  in[u] = timer;
  for (auto h : g[u]) {
    if (h != pr) {
      deep[h] = deep[u] + 1;
      jump[0][h] = u;
      dfs(h, u);
    }
  }
  out[u] = timer;
}
bool isp(int a, int b) { return in[a] <= in[b] && out[b] <= out[a]; }
int lca(int a, int b) {
  if (isp(a, b)) {
    return a;
  }
  if (isp(b, a)) {
    return b;
  }
  for (int k = 18; k >= 0; k--) {
    if (!isp(jump[k][a], b)) {
      a = jump[k][a];
    }
  }
  return jump[0][a];
}
int dist(int a, int b) { return deep[a] + deep[b] - 2 * deep[lca(a, b)]; }
bool comp(int a, int b) { return in[a] < in[b]; }
vector<int> v[MAXN];
int pre[MAXN];
vector<pair<int, int> > sf[MAXN];
int col[MAXN];
int curd[MAXN];
void solve() {
  int k, m;
  cin >> k >> m;
  st.clear();
  for (int i = 1; i <= k; i++) {
    cin >> vs[i] >> s[i];
    vs[i]--;
    st.push_back(vs[i]);
  }
  for (int i = 0; i < m; i++) {
    int u;
    cin >> u;
    u--;
    z[i] = u;
    st.push_back(z[i]);
  }
  sort((st).begin(), (st).end(), comp);
  st.resize(distance(st.begin(), unique((st).begin(), (st).end())));
  vector<int> sts;
  vector<int> ts;
  for (int i = 0; i < (int)(st).size(); i++) {
    int j = st[i];
    ts.push_back(j);
    bool was = false;
    int last = -1;
    while (!sts.empty()) {
      int f = sts.back();
      if (isp(f, j)) {
        int a = (last == -1 ? f : lca(last, j));
        was = true;
        if (a != f) {
          v[f].pop_back();
          v[f].push_back(a);
          ts.push_back(a);
          v[a].push_back(last);
          v[a].push_back(j);
          sts.push_back(a);
          sts.push_back(j);
        } else {
          v[f].push_back(j);
          sts.push_back(j);
        }
        break;
      }
      sts.pop_back();
      last = f;
    }
    if (!was) {
      if (last != -1) {
        was = true;
        int a = lca(last, j);
        ts.push_back(a);
        v[a].push_back(last);
        v[a].push_back(j);
        sts.push_back(a);
        sts.push_back(j);
      } else {
        sts.push_back(j);
      }
    }
  }
  sort((ts).begin(), (ts).end());
  ts.resize(distance(ts.begin(), unique((ts).begin(), (ts).end())));
  for (auto x : ts) {
    sf[x].clear();
    col[x] = 0;
    curd[x] = 0;
  }
  for (int i = 1; i <= k; i++) {
    col[vs[i]] = i;
    curd[vs[i]] = 0;
  }
  for (auto x : ts) {
    for (auto y : v[x]) {
      int len = dist(x, y);
      sf[x].push_back({y, len});
      sf[y].push_back({x, len});
    }
    v[x].clear();
  }
  set<pair<pair<int, int>, pair<int, pair<int, int> > > > kek;
  for (int i = 1; i <= k; i++) {
    for (auto x : sf[vs[i]]) {
      if (col[x.first]) {
        continue;
      }
      kek.insert(
          {{(x.second + s[i] - 1) / s[i], i}, {vs[i], {x.first, x.second}}});
    }
  }
  while (!kek.empty()) {
    auto a = *kek.begin();
    kek.erase(a);
    if (col[a.second.second.first]) {
      continue;
    }
    col[a.second.second.first] = a.first.second;
    curd[a.second.second.first] = curd[a.second.first] + a.second.second.second;
    for (auto x : sf[a.second.second.first]) {
      if (!col[x.first]) {
        int f = curd[a.second.second.first] + x.second;
        kek.insert(
            {{(f + s[a.first.second] - 1) / s[a.first.second], a.first.second},
             {a.second.second.first, x}});
      }
    }
  }
  for (int i = 0; i < m; i++) {
    cout << col[z[i]] << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  for (int i = 1; i <= 18; i++) {
    for (int j = 0; j < n; j++) {
      jump[i][j] = jump[i - 1][jump[i - 1][j]];
    }
  }
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
}