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;
struct bian {
  int nxt, to;
} bi[400040];
int n, m, k, qnum, head[200020], num;
inline void add(int from, int to) {
  bi[++num] = bian{head[from], to};
  head[from] = num;
}
int id[200020], idnum, up[200020][18], dep[200020];
void dfs(int v, int fa) {
  id[v] = ++idnum;
  dep[v] = dep[fa] + 1;
  up[v][0] = fa;
  for (int i = 1; i <= 17; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  for (int i = head[v], u; i; i = bi[i].nxt) {
    u = bi[i].to;
    if (u == fa) continue;
    dfs(u, v);
  }
}
int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int d = dep[x] - dep[y];
  for (int i = 0; i <= 17; ++i)
    if (d >> i & 1) x = up[x][i];
  if (x == y) return x;
  for (int i = 17; ~i; --i)
    if (up[x][i] != up[y][i]) x = up[x][i], y = up[y][i];
  return up[x][0];
}
int qdis(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; }
int s[200020], st[200020], tp, p[200020], cnt, a[200020];
bool vis[200020];
queue<int> q;
vector<int> E[200020];
int fr[200020], kd[200020], ans[200020];
int calc(int x, int y, int p) {
  if (!x || !y) return x + y;
  int d1 = qdis(x, p), d2 = qdis(y, p);
  if ((d1 + s[x] - 1) / s[x] < (d2 + s[y] - 1) / s[y] ||
      ((d1 + s[x] - 1) / s[x] == (d2 + s[y] - 1) / s[y] && kd[x] < kd[y]))
    return x;
  else
    return y;
}
void dfs1(int v) {
  fr[v] = 0;
  if (s[v]) fr[v] = v;
  for (auto u : E[v]) {
    dfs1(u);
    fr[v] = calc(fr[v], fr[u], v);
  }
}
void dfs2(int v, int f) {
  fr[v] = calc(fr[v], f, v);
  ans[v] = kd[fr[v]];
  for (auto u : E[v]) dfs2(u, fr[v]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; ++i) {
    scanf("%d %d", &x, &y);
    add(x, y), add(y, x);
  }
  dfs(1, 0);
  scanf("%d", &qnum);
  while (qnum--) {
    scanf("%d %d", &k, &m);
    for (int i = 1, x; i <= k; ++i) {
      scanf("%d", &x);
      scanf("%d", &s[x]);
      kd[x] = i;
      if (!vis[x]) vis[x] = 1, q.push(x), p[++cnt] = x;
    }
    for (int i = 1, x; i <= m; ++i) {
      scanf("%d", &x);
      a[i] = x;
      if (!vis[x]) vis[x] = 1, q.push(x), p[++cnt] = x;
    }
    if (!vis[1]) vis[1] = 1, q.push(1), p[++cnt] = 1;
    sort(p + 1, p + 1 + cnt, [&](int x, int y) { return id[x] < id[y]; });
    st[tp = 1] = 1;
    for (int i = 2; i <= cnt; ++i) {
      int l = lca(st[tp], p[i]);
      while (dep[l] < dep[st[tp]]) {
        if (dep[l] <= dep[st[tp - 1]]) {
          E[st[tp - 1]].emplace_back(st[tp]);
          --tp;
        } else {
          E[l].emplace_back(st[tp]);
          st[tp] = l;
          q.push(l);
        }
      }
      st[++tp] = p[i];
    }
    while (tp > 1) E[st[tp - 1]].emplace_back(st[tp]), --tp;
    dfs1(1);
    dfs2(1, 0);
    for (int i = 1; i <= m; ++i) printf("%d ", ans[a[i]]);
    printf("\n");
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      E[x].clear();
      vis[x] = 0, s[x] = 0;
    }
    cnt = 0;
  }
  return 0;
}