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
#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-', ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
const int N = 200005;
int n;
vector<int> e[N];
int fa[N][20], d[N], I[N], O[N];
void dfs(int x, int pre, int dep) {
  static int Time = 0;
  I[x] = ++Time, d[x] = dep, fa[x][0] = pre;
  for (int i = (1); i <= (19); i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (auto y : e[x])
    if (y != pre) dfs(y, x, dep + 1);
  O[x] = Time;
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = (19); i >= (0); i--)
    if (d[x] - (1 << i) >= d[y]) x = fa[x][i];
  if (x == y) return x;
  for (int i = (19); i >= (0); i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
int dis(int x, int y) { return d[x] + d[y] - d[lca(x, y)] * 2; }
int k, m;
vector<int> ve[N];
void pve(int x, int y) { ve[x].push_back(y), ve[y].push_back(x); }
int res[N];
struct P {
  int x, v;
  long long t;
  P() {}
  P(int _x, int _v, long long _t) { x = _x, v = _v, t = _t; }
  friend bool operator<(P a, P b) { return a.t > b.t; }
};
void Solve() {
  k = read(), m = read();
  vector<int> v(k + 1), s(k + 1), u(m), id(1, 1);
  for (int i = (1); i <= (k); i++)
    v[i] = read(), s[i] = read(), id.push_back(v[i]);
  for (int i = (0); i <= (m - 1); i++) u[i] = read(), id.push_back(u[i]);
  sort(id.begin(), id.end());
  id.resize(unique(id.begin(), id.end()) - id.begin());
  sort(id.begin(), id.end(), [&](int a, int b) -> bool { return I[a] < I[b]; });
  static int st[N];
  int top = 0;
  vector<int> tid = id;
  for (auto i : id) {
    ve[i].clear();
    if (top > 0) {
      int f = lca(i, st[top]);
      while (d[st[top]] > d[f]) {
        if (d[st[top - 1]] >= d[f])
          pve(st[top], st[top - 1]), top--;
        else {
          tid.push_back(f), ve[f].clear();
          pve(st[top], f), st[top] = f;
        }
      }
    }
    st[++top] = i;
  }
  while (top > 1) pve(st[top], st[top - 1]), top--;
  for (auto i : tid) res[i] = 0;
  priority_queue<P> Q;
  for (int i = (1); i <= (k); i++) Q.push(P(v[i], i, i));
  while (!Q.empty()) {
    P p = Q.top();
    Q.pop();
    if (res[p.x]) continue;
    res[p.x] = p.v;
    for (auto y : ve[p.x])
      if (!res[y])
        Q.push(
            P(y, p.v,
              (long long)((dis(v[p.v], y) + s[p.v] - 1) / s[p.v]) * k + p.v));
  }
  for (int i = (0); i <= (m - 1); i++) printf("%d ", res[u[i]]);
  puts("");
}
int main() {
  n = read();
  for (int i = (1); i <= (n - 1); i++) {
    int x = read(), y = read();
    e[x].push_back(y), e[y].push_back(x);
  }
  dfs(1, 0, 0);
  int q = read();
  while (q--) Solve();
  return 0;
}
