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 = 200005;
int ne[N << 1], fi[N], zz[N << 1], sd[N], fa[N][20], in[N], Stack[N], c[N << 1],
    a[N], b[N], cnt1, cnt2, s[N], vis[N], x, y, tot, n, q, out[N];
vector<int> g, V[N];
struct Ask {
  int val, from, to;
  inline Ask(int _val = 0, int _from = 0, int _to = 0) {
    val = _val;
    from = _from;
    to = _to;
  }
  inline bool operator<(const Ask &b) const {
    return val < b.val || (val == b.val && from < b.from) ||
           (val == b.val && from == b.from && to < b.to);
  }
};
set<Ask> S;
inline int cmp(int x, int y) { return in[x] < in[y]; }
inline int pd(int x, int y) {
  if (!x) return 1;
  return in[x] <= in[y] && out[x] >= out[y];
}
inline void jb(int x, int y) {
  ne[++tot] = fi[x];
  fi[x] = tot;
  zz[tot] = y;
}
inline int lca(int x, int y) {
  if (x == y) return x;
  if (sd[x] < sd[y]) swap(x, y);
  for (int i = 19; i >= 0; i--)
    if (!pd(fa[x][i], y)) x = fa[x][i];
  return fa[x][0];
}
inline int dist(int x, int y) { return sd[x] + sd[y] - 2 * sd[lca(x, y)]; }
void dfs(int x, int y, int z) {
  in[x] = ++tot;
  fa[x][0] = y;
  sd[x] = z;
  for (int i = fi[x]; i; i = ne[i])
    if (zz[i] != y) dfs(zz[i], x, z + 1);
  out[x] = tot;
}
inline void upd(int x, int y) {
  vis[x] = y;
  g.push_back(x);
  for (int i : V[x])
    if (!vis[i]) S.insert(Ask((dist(a[y], i) - 1) / s[y] + 1, y, i));
}
inline void Insert(int x) {
  if (tot <= 1) {
    Stack[++tot] = x;
    return;
  }
  int l = lca(Stack[tot], x);
  if (l == Stack[tot]) {
    Stack[++tot] = x;
    return;
  }
  while (tot > 1 && in[l] <= in[Stack[tot - 1]]) {
    V[Stack[tot - 1]].push_back(Stack[tot]);
    V[Stack[tot]].push_back(Stack[tot - 1]);
    tot--;
  }
  if (l != Stack[tot]) {
    V[l].push_back(Stack[tot]);
    V[Stack[tot]].push_back(l);
    Stack[tot] = l;
  }
  Stack[++tot] = x;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    jb(x, y);
    jb(y, x);
  }
  tot = 0;
  dfs(1, 0, 0);
  for (int i = 1; i <= 19; i++)
    for (int j = 1; j <= n; j++) fa[j][i] = fa[fa[j][i - 1]][i - 1];
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &cnt1, &cnt2);
    int cnt = 0;
    tot = 0;
    for (int i = 1; i <= cnt1; i++)
      scanf("%d%d", &a[i], &s[i]), c[++cnt] = a[i];
    for (int i = 1; i <= cnt2; i++) scanf("%d", &b[i]), c[++cnt] = b[i];
    sort(c + 1, c + cnt + 1, cmp);
    Insert(1);
    for (int i = 1; i <= cnt; i++)
      if (c[i] != c[i - 1] && c[i] != 1) Insert(c[i]);
    for (int i = tot - 1; i; i--)
      V[Stack[i]].push_back(Stack[i + 1]), V[Stack[i + 1]].push_back(Stack[i]);
    for (int i = 1; i <= cnt1; i++) upd(a[i], i);
    while (S.size()) {
      Ask iter = *S.begin();
      S.erase(S.begin());
      if (vis[iter.to]) continue;
      upd(iter.to, iter.from);
    }
    for (int i = 1; i <= cnt2; i++) printf("%d ", vis[b[i]]);
    puts("");
    for (int i : g) vis[i] = 0, V[i].clear();
    g.clear();
  }
  return 0;
}