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;
namespace RA {
int r(int p) { return 1ll * rand() * rand() % p; }
int r(int L, int R) { return r(R - L + 1) + L; }
}  // namespace RA
const int N = 5e5 + 5;
int n;
vector<int> g[N];
vector<pair<int, int> > vt[N];
int dep[N], fa[N][20];
int dfn[N], totdfn;
void dfs(int u, int p) {
  dep[u] = dep[p] + 1;
  fa[u][0] = p;
  dfn[u] = ++totdfn;
  for (int j = (1); j <= (19); ++j) {
    fa[u][j] = fa[fa[u][j - 1]][j - 1];
    if (fa[u][j] == 0) break;
  }
  for (int v : g[u])
    if (v != p) dfs(v, u);
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int j = (19); j >= (0); --j)
    if (dep[x] - (1 << j) >= dep[y]) x = fa[x][j];
  if (x == y) return x;
  for (int j = (19); j >= (0); --j)
    if (fa[x][j] != fa[y][j]) x = fa[x][j], y = fa[y][j];
  return fa[x][0];
}
int distance(int x, int y) { return dep[x] + dep[y] - dep[lca(x, y)] * 2; }
void VT_addpath(int u, int v, int w) {
  ;
  ;
  vt[u].push_back({v, w});
  vt[v].push_back({u, w});
}
int s[N], tp;
void build_VT(vector<int>& V) {
  for (int u : V) vt[u].clear();
  sort(V.begin(), V.end(), [](int x, int y) { return dfn[x] < dfn[y]; });
  V.resize(unique(V.begin(), V.end()) - V.begin());
  tp = 0;
  s[++tp] = lca(V.front(), V.back());
  vector<int> Vt;
  for (int u : V) {
    int z = lca(s[tp], u);
    ;
    ;
    if (u == z) continue;
    while (tp > 1 && dep[s[tp - 1]] >= dep[z])
      VT_addpath(s[tp], s[tp - 1], distance(s[tp], s[tp - 1])), --tp;
    if (s[tp] != z)
      VT_addpath(s[tp], z, distance(s[tp], z)), s[tp] = z, Vt.push_back(z);
    s[++tp] = u;
  }
  while (tp > 1) VT_addpath(s[tp], s[tp - 1], distance(s[tp], s[tp - 1])), --tp;
  V.push_back(s[1]);
  V.insert(V.end(), Vt.begin(), Vt.end());
  sort(V.begin(), V.end(), [](int x, int y) { return dfn[x] < dfn[y]; });
  V.resize(unique(V.begin(), V.end()) - V.begin());
}
int k, m;
struct virus {
  int v, s, dist, id;
  void reset() { dist = id = 1e9, s = 1, v = 0; }
  virus() { reset(); }
  int tim() const { return dist / s + !!(dist % s); }
  bool operator<(virus x) const {
    return (tim() != x.tim()) ? tim() > x.tim() : id > x.id;
  }
  bool operator>(virus x) const {
    return (tim() != x.tim()) ? tim() < x.tim() : id < x.id;
  }
  bool operator>=(virus x) const {
    return (tim() != x.tim()) ? tim() < x.tim() : id <= x.id;
  }
  virus walkto(int u, int d) const {
    virus res = *this;
    res.dist += d, res.v = u;
    return res;
  }
} vss[N];
int a[N];
priority_queue<virus> q;
virus t[N];
void go() {
  scanf("%d%d", &k, &m);
  for (int i = (1); i <= (k); ++i)
    scanf("%d%d", &vss[i].v, &vss[i].s), vss[i].dist = 0, vss[i].id = i;
  for (int i = (1); i <= (m); ++i) scanf("%d", &a[i]);
  vector<int> V;
  for (int i = (1); i <= (k); ++i) V.push_back(vss[i].v);
  for (int i = (1); i <= (m); ++i) V.push_back(a[i]);
  build_VT(V);
  for (int v : V) t[v].reset();
  for (int i = (1); i <= (k); ++i) q.push(vss[i]), t[vss[i].v] = vss[i];
  while (!q.empty()) {
    virus vr = q.top();
    q.pop();
    int u = vr.v;
    ;
    ;
    if (t[u] > vr) continue;
    for (pair<int, int> p : vt[u]) {
      int v = p.first, w = p.second;
      ;
      ;
      virus nv = vr.walkto(v, w);
      if (t[v] >= nv) continue;
      t[v] = nv;
      q.push(nv);
    }
  }
  for (int i = (1); i <= (m); ++i) printf("%d%c", t[a[i]].id, " \n"[i == m]);
  for (int v : V) vt[v].clear();
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  int q;
  scanf("%d", &q);
  for (int i = (1); i <= (q); ++i) go();
  return 0;
}