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 = (int)2e5 + 50, INF = (int)1e9 + 50;
const int LOG_N = 18;
int n;
int root = 0;
vector<int> G[N];
int parent[LOG_N][N];
int depth[N], dfn[N], T = 0;
void dfs(int v, int p, int d) {
  dfn[v] = T++;
  parent[0][v] = p;
  depth[v] = d;
  for (int nxt : G[v]) {
    if (nxt != p) dfs(nxt, v, d + 1);
  }
}
void init(int V) {
  dfs(root, -1, 0);
  for (int k = 0; k + 1 < LOG_N; k++) {
    for (int v = 0; v < V; v++) {
      if (parent[k][v] < 0)
        parent[k + 1][v] = -1;
      else
        parent[k + 1][v] = parent[k][parent[k][v]];
    }
  }
}
int lca(int u, int v) {
  if (depth[u] > depth[v]) {
    int tmp = u;
    u = v;
    v = tmp;
  }
  for (int k = 0; k < LOG_N; k++) {
    if ((depth[v] - depth[u]) >> k & 1) {
      v = parent[k][v];
    }
  }
  if (u == v) return u;
  for (int k = LOG_N - 1; k >= 0; k--) {
    if (parent[k][u] != parent[k][v]) {
      u = parent[k][u];
      v = parent[k][v];
    }
  }
  return parent[0][u];
}
int min_dis(int u, int v) {
  int ca = lca(u, v);
  return depth[u] + depth[v] - 2 * depth[ca];
}
struct edge {
  int to, cost;
};
int stk[N];
vector<int> vir;
vector<edge> V[N];
void add_edge(int u, int v) {
  int cost = min_dis(u, v);
  V[u].push_back({v, cost});
  V[v].push_back({u, cost});
}
void build(vector<int> vec) {
  for (auto u : vir) V[u].clear();
  vir = vec;
  sort(vec.begin(), vec.end(), [](int u, int v) { return dfn[u] < dfn[v]; });
  int sz = 0;
  stk[++sz] = vec[0];
  for (auto u : vec) {
    int x = lca(u, stk[sz]);
    vir.push_back(x);
    if (x != stk[sz]) {
      while (sz >= 2 && depth[stk[sz - 1]] >= depth[x]) {
        add_edge(stk[sz - 1], stk[sz]);
        sz--;
      }
      if (x != stk[sz]) {
        add_edge(x, stk[sz]);
        stk[sz] = x;
      }
    }
    if (u != stk[sz]) stk[++sz] = u;
  }
  for (int i = 1; i <= sz - 1; i++) add_edge(stk[i], stk[i + 1]);
}
int v[N], s[N], u[N];
struct Cost {
  int sp, cost;
  bool operator<(const Cost &o) const {
    if (sp == -1) return false;
    if (o.sp == -1) return true;
    int bk1 = (cost + s[sp] - 1) / s[sp],
        bk2 = (o.cost + s[o.sp] - 1) / s[o.sp];
    return bk1 != bk2 ? bk1 < bk2 : (sp != o.sp ? sp < o.sp : cost < o.cost);
  }
};
Cost dis[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  init(n);
  int q;
  cin >> q;
  while (q--) {
    int k, m;
    cin >> k >> m;
    vector<int> ns;
    for (int i = 0; i < k; i++) {
      cin >> v[i] >> s[i];
      v[i]--;
      ns.push_back(v[i]);
    }
    for (int i = 0; i < m; i++) {
      cin >> u[i];
      u[i]--;
      ns.push_back(u[i]);
    }
    sort(ns.begin(), ns.end());
    ns.resize(unique(ns.begin(), ns.end()) - ns.begin());
    build(ns);
    for (auto v : vir) dis[v] = {-1, -1};
    priority_queue<pair<Cost, int>, vector<pair<Cost, int> >,
                   greater<pair<Cost, int> > >
        pque;
    for (int i = 0; i < k; i++) {
      dis[v[i]] = {i, 0};
      pque.push({dis[v[i]], v[i]});
    }
    while (!pque.empty()) {
      auto p = pque.top();
      pque.pop();
      if (dis[p.second] < p.first) continue;
      Cost c = p.first;
      int v = p.second;
      for (auto e : V[v]) {
        Cost nxtc = {c.sp, c.cost + e.cost};
        if (nxtc < dis[e.to]) {
          dis[e.to] = nxtc;
          pque.push({nxtc, e.to});
        }
      }
    }
    for (int i = 0; i < m; i++) {
      cout << dis[u[i]].sp + 1 << " ";
    }
    cout << "\n";
  }
}