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;
namespace FGF {
int n, m, K, Q;
const int N = 2e5 + 5;
vector<int> g[N], G[N];
int dfn[N], top[N], dep[N], fa[N], num, tot, siz[N], son[N], a[N], s[N], b[N],
    q[N], col[N], dis[N];
bool vis[N];
void dfs1(int u, int f) {
  fa[u] = f, dep[u] = dep[f] + 1, siz[u] = 1;
  for (auto v : g[u]) {
    if (v == f) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int tp) {
  top[u] = tp, dfn[u] = ++num;
  if (son[u]) dfs2(son[u], tp);
  for (auto v : g[u])
    if (v != son[u] && v != fa[u]) dfs2(v, v);
}
int getlca(int u, int v) {
  while (top[u] != top[v])
    dep[top[u]] > dep[top[v]] ? u = fa[top[u]] : v = fa[top[v]];
  return dep[u] < dep[v] ? u : v;
}
int getdis(int u, int v) { return dep[u] + dep[v] - 2 * dep[getlca(u, v)]; }
bool cmp(int u, int v) { return dfn[u] < dfn[v]; }
int st[N], tp;
struct Node {
  int v, w, c;
  Node() {}
  Node(int _v, int _w, int _c) : v(_v), w(_w), c(_c){};
};
bool operator<(Node x, Node y) { return x.w ^ y.w ? x.w > y.w : x.c > y.c; }
void add(int u, int v) { G[u].push_back(v), G[v].push_back(u); }
void dij() {
  priority_queue<Node> q;
  for (int i = 1; i <= K; i++)
    q.push(Node(b[i], 0, i)), dis[b[i]] = 0, col[b[i]] = i;
  while (q.size()) {
    int u = q.top().v;
    q.pop();
    if (vis[u]) continue;
    vis[u] = 1;
    for (auto v : G[u]) {
      int d = (getdis(b[col[u]], v) + s[b[col[u]]] - 1) / s[b[col[u]]];
      if (!vis[v] && (dis[v] > d || (dis[v] == d && col[u] < col[v])))
        col[v] = col[u], dis[v] = d, q.push(Node(v, d, col[v]));
    }
    G[u].clear();
  }
}
void inser(int x) {
  if (x == 1) return;
  dis[x] = 0x3f3f3f3f, vis[x] = 0, col[x] = 0x3f3f3f3f;
  int lca = getlca(x, st[tp]);
  while (tp > 1 && dfn[st[tp - 1]] >= dfn[lca]) add(st[tp - 1], st[tp]), tp--;
  if (lca != st[tp])
    add(lca, st[tp]), st[tp] = lca, dis[lca] = 0x3f3f3f3f, vis[lca] = 0,
                      col[x] = 0x3f3f3f3f;
  st[++tp] = x;
}
void work() {
  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);
  }
  dfs1(1, 0), dfs2(1, 1);
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &K, &m);
    for (int i = 1, x; i <= K; i++)
      scanf("%d%d", &b[i], &x), s[b[i]] = x, a[i] = b[i];
    for (int i = 1; i <= m; i++) scanf("%d", &q[i]), a[i + K] = q[i];
    sort(a + 1, a + K + m + 1, cmp);
    int tot = unique(a + 1, a + K + m + 1) - a - 1;
    st[tp = 1] = 1, dis[1] = 0x3f3f3f3f, vis[1] = 0, col[1] = 0x3f3f3f3f;
    for (int i = 1; i <= tot; i++) inser(a[i]);
    while (tp > 1) add(st[tp - 1], st[tp]), tp--;
    dij();
    for (int i = 1; i <= m; i++) printf("%d ", col[q[i]]);
    puts(" ");
  }
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}