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 = 2e5 + 9;
const long long mod = 998244353;
vector<int> g[N], nt[N];
vector<pair<int, int> > nod, fn;
int fa[N][20], dfn[N], in, t[N], p[N], s[N], sz[N], dep[N], id[N], top, stk[N];
pair<int, int> dp[N];
void dfs(int u) {
  dfn[u] = ++in;
  sz[u] = 1;
  for (auto v : g[u]) {
    if (v == fa[u][0]) continue;
    fa[v][0] = u;
    dep[v] = dep[u] + 1;
    dfs(v);
    sz[u] += sz[v];
  }
}
int LCA(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int d = dep[u] - dep[v];
  for (int i = (20) - 1; i >= (0); --i)
    if (d & (1 << i)) u = fa[u][i];
  if (u == v) return u;
  for (int i = (20) - 1; i >= (0); --i)
    if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];
  return fa[u][0];
}
bool chk(int u, int v) { return dfn[u] + sz[u] - 1 >= dfn[v]; }
int dis(int u, int v) { return dep[u] + dep[v] - 2 * dep[LCA(u, v)]; }
pair<int, int> cal(int u, int x) {
  int d = dis(u, t[x]);
  int cu = d / s[x];
  if (d % s[x]) ++cu;
  return {cu, x};
}
void work(int u) {
  if (!id[u])
    dp[u] = {1e9, 0};
  else
    dp[u] = {0, id[u]};
  for (auto v : nt[u]) {
    work(v);
    if (dp[v].second) dp[u] = min(dp[u], cal(u, dp[v].second));
  }
}
void work2(int u) {
  for (auto v : nt[u]) {
    dp[v] = min(dp[v], cal(v, dp[u].second));
    work2(v);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (1); i < (n); ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  for (int i = (1); i < (20); ++i)
    for (int j = (1); j < (n + 1); ++j) fa[j][i] = fa[fa[j][i - 1]][i - 1];
  int q;
  scanf("%d", &q);
  while (q--) {
    nod.clear();
    int k, m;
    scanf("%d%d", &k, &m);
    for (int i = (1); i < (k + 1); ++i)
      scanf("%d%d", &t[i], &s[i]), nod.push_back({dfn[t[i]], t[i]}),
          id[t[i]] = i;
    for (int i = (1); i < (m + 1); ++i)
      scanf("%d", &p[i]), nod.push_back({dfn[p[i]], p[i]});
    sort(nod.begin(), nod.end());
    fn = nod;
    for (int i = (0); i < (nod.size() - 1); ++i) {
      int lca = LCA(nod[i].second, nod[i + 1].second);
      fn.push_back({dfn[lca], lca});
    }
    fn.push_back({1, 1});
    sort(fn.begin(), fn.end());
    top = 0;
    stk[top++] = fn[0].second;
    for (int i = (1); i < (fn.size()); ++i) {
      if (fn[i].second == fn[i - 1].second) continue;
      while (!chk(stk[top - 1], fn[i].second)) --top;
      nt[stk[top - 1]].push_back(fn[i].second);
      stk[top++] = fn[i].second;
    }
    work(1);
    work2(1);
    for (int i = (1); i < (k + 1); ++i) id[t[i]] = 0;
    for (int i = (0); i < (fn.size()); ++i) nt[fn[i].second].clear();
    for (int i = (1); i < (m + 1); ++i) printf("%d ", dp[p[i]].second);
    printf("\n");
  }
}