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;
int read() {
  int x = 0, f = 1, ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar();
  return x * f;
}
int dfn[200005], tot = 0, dep[200005], fa[200005], top[200005], son[200005],
                 size[200005];
int n, Q, x[200005], s[200005], qr[200005], S = 200003;
int inf[200005], dis[200005];
int cal(int x, int y) { return x / y + (x % y > 0); }
struct node {
  int dis, v, s, id;
  bool operator<(const node &b) const {
    return cal(dis, s) == cal(b.dis, b.s) ? id > b.id
                                          : cal(dis, s) > cal(b.dis, b.s);
  }
};
priority_queue<node> q;
bool vis[200005];
struct EDGE {
  struct edge {
    int to, w, nxt;
  } e[200005 << 1];
  int head[200005], k = 0;
  void init() {
    memset(head, -1, sizeof head);
    k = 0;
  }
  void add(int u, int v, int w) {
    e[k] = {v, w, head[u]};
    head[u] = k++;
  }
  void dfs1(int u) {
    size[u] = 1, dfn[u] = ++tot;
    for (int i = head[u], v; ~i; i = e[i].nxt) {
      v = e[i].to;
      if (v == fa[u]) continue;
      fa[v] = u, dep[v] = dep[u] + 1;
      dfs1(v);
      size[u] += size[v];
      if (size[v] > size[son[u]]) son[u] = v;
    }
  }
  void dfs2(int u, int tp) {
    top[u] = tp;
    if (son[u]) dfs2(son[u], tp);
    for (int i = head[u], v; ~i; i = e[i].nxt) {
      v = e[i].to;
      if (v != fa[u] && v != son[u]) dfs2(v, v);
    }
  }
  int LCA(int u, int v) {
    while (top[u] != top[v]) {
      if (dep[top[u]] > dep[top[v]]) swap(u, v);
      v = fa[top[v]];
    }
    if (dep[u] > dep[v]) return v;
    return u;
  }
  void dij(int st) {
    for (int i = 1; i <= st; i++) dis[x[i]] = 0, q.push({0, x[i], s[i], i});
    while (q.size()) {
      node now = q.top();
      q.pop();
      if (cal(now.dis, now.s) != dis[now.v] || vis[now.v]) continue;
      register int u = now.v, v;
      vis[u] = true;
      for (int i = head[u], v; ~i; i = e[i].nxt) {
        register int tmp_d = cal(now.dis + e[i].w, now.s);
        v = e[i].to;
        if (tmp_d < dis[v] || (tmp_d == dis[v] && now.id < inf[v])) {
          dis[v] = tmp_d, inf[v] = now.id;
          q.push({now.dis + e[i].w, v, s[now.id], now.id});
        }
      }
    }
  }
} et, e;
int stc[200005], tp = 0;
bool cmp_dfn(int a, int b) { return dfn[a] < dfn[b]; }
bool in[200005];
signed main() {
  n = read();
  et.init();
  e.init();
  for (int i = 1, u, v; i < n; i++)
    u = read(), v = read(), et.add(u, v, 1), et.add(v, u, 1);
  et.dfs1(1);
  et.dfs2(1, 1);
  Q = read();
  memset(dis, 0x3f, sizeof dis);
  while (Q--) {
    register int k = read(), m = read();
    tp = 0;
    for (int i = 1; i <= k; i++)
      x[i] = read(), s[i] = read(), stc[++tp] = x[i], in[x[i]] = true,
      inf[x[i]] = i;
    for (int i = 1; i <= m; i++) {
      qr[i] = read();
      if (!in[qr[i]]) stc[++tp] = qr[i], in[qr[i]] = true;
    }
    sort(stc + 1, stc + 1 + tp, cmp_dfn);
    for (int i = 1, tmp = tp; i < tmp; i++) {
      register int lca = et.LCA(stc[i], stc[i + 1]);
      if (!in[lca]) stc[++tp] = lca, in[lca] = true;
    }
    sort(stc + 1, stc + 1 + tp, cmp_dfn);
    for (int i = 2; i <= tp; i++) {
      register int lca = et.LCA(stc[i - 1], stc[i]), d = dep[stc[i]] - dep[lca];
      e.add(stc[i], lca, d), e.add(lca, stc[i], d);
    }
    e.dij(k);
    for (int i = 1; i <= m; i++) printf("%d ", inf[qr[i]]);
    puts("");
    for (int i = 1; i <= tp; i++)
      in[stc[i]] = inf[stc[i]] = vis[stc[i]] = 0, dis[stc[i]] = 0x3f3f3f3f,
      e.head[stc[i]] = -1;
    e.k = 0;
  }
  return 0;
}