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 = 4e5 + 5;
int n, Q, k, m, sz[N], dfn[N], son[N];
int fa[N], top[N], dep[N], ids, p[N];
basic_string<int> G[N], T[N];
struct Nod {
  int c, v, d, t;
  bool operator<(const Nod &x) const {
    if (t != x.t) return t > x.t;
    if (c != x.c) return c > x.c;
    return d < x.d;
  }
} ans[N], I;
using pni = pair<Nod, int>;
priority_queue<pni> q;
void dfs1(int u) {
  sz[u] = 1, dep[u] = dep[fa[u]] + 1;
  for (int v : G[u])
    if (v != fa[u]) {
      fa[v] = u, dfs1(v), sz[u] += sz[v];
      if (sz[v] > sz[son[u]]) son[u] = v;
    }
}
void dfs2(int u, int tp) {
  top[u] = tp, dfn[u] = ++ids;
  if (son[u]) dfs2(son[u], tp);
  for (int v : G[u])
    if (v && !dfn[v]) dfs2(v, v);
}
int LCA(int x, int y) {
  for (; top[x] != top[y]; x = fa[top[x]])
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
  return dep[x] > dep[y] ? y : x;
}
inline int dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[LCA(x, y)]; }
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
void build_V() {
  static int a[N * 2], cnt;
  a[cnt = 1] = 1;
  for (int i = 1; i <= k; ++i) {
    scanf("%d", a + (++cnt)), ans[a[cnt]].c = i;
    scanf("%d", &ans[a[cnt]].v);
    ans[a[cnt]].t = 0, q.push({ans[a[cnt]], a[cnt]});
  }
  for (int i = 1; i <= m; ++i) scanf("%d", p + i), a[++cnt] = p[i];
  sort(a + 1, a + cnt + 1, cmp);
  for (int i = 1; i < cnt; ++i) a[cnt + i] = LCA(a[i], a[i + 1]);
  sort(a + 1, a + cnt * 2, cmp), cnt = unique(a + 1, a + cnt * 2) - a - 1;
  for (int i = 2, t; i <= cnt; ++i)
    T[a[i]] += (t = LCA(a[i], a[i - 1])), T[t] += a[i];
}
Nod exbd(int u, int v) {
  int d = dis(u, v);
  Nod tmp = ans[u];
  if (tmp.d >= d) {
    tmp.d -= d;
    return tmp;
  }
  d -= tmp.d;
  int t = (d - 1) / tmp.v + 1;
  tmp.d = t * tmp.v - d, tmp.t += t;
  return tmp;
}
void dij() {
  while (!q.empty()) {
    pni u = q.top();
    q.pop();
    if (u.first < ans[u.second]) continue;
    for (int v : T[u.second]) {
      Nod vv = exbd(u.second, v);
      if (ans[v] < vv) ans[v] = vv, q.push({ans[v], v});
    }
  }
}
void clear(int u, int fa) {
  for (int v : T[u])
    if (v != fa) clear(v, u);
  T[u] = {}, ans[u] = I;
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; ++i) scanf("%d%d", &x, &y), G[x] += y, G[y] += x;
  scanf("%d", &Q), dfs1(1), dfs2(1, 1), I.c = I.t = n + 1;
  for (int i = 1; i <= n; ++i) ans[i] = I;
  for (int j = 1; j <= Q; ++j) {
    scanf("%d%d", &k, &m), build_V(), dij();
    for (int i = 1; i <= m; ++i) printf("%d ", ans[p[i]].c);
    puts(""), clear(1, 0);
  }
  return 0;
}