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>
const int N = 400200;
const int Q = 400200;
using namespace std;
int n;
int tim;
int dip[N];
int tin[N];
int tout[N];
int f[N][20];
vector<int> v[N];
void dfs(int x, int p) {
  tin[x] = ++tim;
  f[x][0] = p;
  for (int i = 1; i < 20; i++) f[x][i] = f[f[x][i - 1]][i - 1];
  for (int y : v[x]) {
    if (y == p) continue;
    dip[y] = dip[x] + 1;
    dfs(y, x);
  }
  tout[x] = ++tim;
}
bool isp(int x, int y) { return tin[x] <= tin[y] && tout[x] >= tout[y]; }
int lca(int x, int y) {
  if (isp(x, y)) return x;
  if (isp(y, x)) return y;
  for (int i = 19; i >= 0; i--)
    if (!isp(f[x][i], y)) x = f[x][i];
  return f[x][0];
}
int V[N];
int s[N];
int a[N];
int id[N];
int inv[N];
int spec[N];
vector<pair<int, int> > g[N];
int get_len(int x, int y) { return dip[x] + dip[y] - 2 * dip[lca(x, y)]; }
void make(set<pair<pair<int, int>, int> > &S, int x) {
  int i = a[x];
  for (auto p : g[x]) {
    if (a[p.first]) {
      continue;
    }
    S.insert({{(get_len(V[i], inv[p.first]) - 1) / s[i] + 1, i}, p.first});
  }
}
pair<int, int> T[N];
void upd_(int x, int y) {
  while (x < N) {
    T[x] = max(T[x], make_pair(tin[y], y));
    x += x & -x;
  }
}
void cler(int x) {
  while (x < N) {
    T[x] = {0, 0};
    x += x & -x;
  }
}
pair<int, int> get_(int x) {
  pair<int, int> res = {0, 0};
  while (x > 0) {
    res = max(res, T[x]);
    x -= x & -x;
  }
  return res;
}
void solve() {
  int k, m;
  scanf("%d%d", &k, &m);
  vector<int> v;
  for (int i = 1; i <= k; i++) {
    scanf("%d%d", V + i, s + i);
    v.push_back(V[i]);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", spec + i);
    v.push_back(spec[i]);
  }
  sort(v.begin(), v.end(), [&](int x, int y) { return tin[x] < tin[y]; });
  int len = v.size();
  for (int i = 1; i < len; i++) {
    if (v[i] == v[i - 1]) {
      continue;
    }
    v.push_back(lca(v[i - 1], v[i]));
  }
  sort(v.begin(), v.end(), [&](int x, int y) {
    return make_pair(tin[x], x) < make_pair(tin[y], y);
  });
  vector<int> nv = {v[0]};
  for (int i = 1; i < v.size(); i++)
    if (v[i] != v[i - 1]) nv.push_back(v[i]);
  v = nv;
  for (int i = 1; i <= v.size(); i++) g[i].clear();
  for (int i = 0, len; i < v.size(); i++) {
    auto q = get_(N - tin[v[i]]);
    id[v[i]] = i + 1;
    inv[i + 1] = v[i];
    if (q.second > 0) {
      len = dip[v[i]] - dip[q.second];
      g[id[q.second]].push_back({i + 1, len});
      g[i + 1].push_back({id[q.second], len});
    }
    upd_(N - tout[v[i]], v[i]);
  }
  for (int i = 0; i < v.size(); i++) {
    cler(N - tout[v[i]]);
  }
  for (int i = 1; i <= v.size(); i++) a[i] = 0;
  set<pair<pair<int, int>, int> > S;
  for (int i = 1; i <= k; i++) {
    int x = V[i];
    a[id[x]] = i;
  }
  for (int i = 1; i <= k; i++) {
    make(S, id[V[i]]);
  }
  while (!S.empty()) {
    auto shit = *S.begin();
    S.erase(S.begin());
    if (a[shit.second]) {
      continue;
    }
    a[shit.second] = shit.first.second;
    make(S, shit.second);
  }
  for (int i = 1; i <= m; i++) {
    printf("%d ", a[id[spec[i]]]);
  }
  printf("\n");
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    solve();
  }
}