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 INF = 1000000000;
int read() {
  int X = 0;
  char ch = 0;
  while (ch < 48 || ch > 57) ch = getchar();
  while (ch >= 48 && ch <= 57) X = X * 10 + (ch ^ 48), ch = getchar();
  return X;
}
vector<int> g[200500];
struct Node {
  int f, tf, p;
} b[200500];
int siz[200500], dep[200500], in[200500], out[200500], tim;
void pfs1(int u) {
  siz[u] = 1;
  for (int i = 0, v; i < g[u].size(); i++)
    if (!siz[v = g[u][i]]) {
      dep[v] = dep[b[v].f = u] + 1;
      pfs1(v);
      siz[u] += siz[v];
      if (siz[v] > siz[b[u].p]) b[u].p = v;
    }
}
void pfs2(int u, int tf) {
  b[u].tf = tf;
  in[u] = ++tim;
  if (b[u].p) pfs2(b[u].p, tf);
  for (int i = 0; i < g[u].size(); i++)
    if (!b[g[u][i]].tf) pfs2(g[u][i], g[u][i]);
  out[u] = tim;
}
int lca(int u, int v) {
  while (b[u].tf != b[v].tf) {
    if (dep[b[u].tf] < dep[b[v].tf]) swap(u, v);
    u = b[b[u].tf].f;
  }
  return dep[u] < dep[v] ? u : v;
}
bool cmp(int A, int B) { return in[A] < in[B]; }
struct Line {
  int t, l, nxt;
} l[200500 << 1];
int fir[200500], tl;
void adl(int u, int v, int len) {
  l[++tl] = (Line){v, len, fir[u]};
  fir[u] = tl;
  l[++tl] = (Line){u, len, fir[v]};
  fir[v] = tl;
}
int p0[200500], p1[200500], p[200500], stk[200500], vis[200500], ef, tn,
    w[200500], dis[200500], dis2[200500], col[200500];
struct Data {
  int d, c, p;
  bool operator<(const Data &B) const { return d == B.d ? c > B.c : d > B.d; }
};
priority_queue<Data> q;
void adp(const int u) {
  if (vis[u] < ef) vis[p[++tn] = u] = ef;
}
void solve() {
  vis[p[tn = 1] = 1] = ++ef;
  int m0 = read(), m1 = read(), top;
  for (int i = 1; i <= m0; i++) {
    adp(p0[i] = read());
    w[i] = read();
  }
  for (int i = 1; i <= m1; i++) adp(p1[i] = read());
  sort(p + 1, p + tn + 1, cmp);
  int sav = tn;
  for (int i = 1; i < sav; i++) adp(lca(p[i], p[i + 1]));
  sort(p + 1, p + tn + 1, cmp);
  stk[top = 1] = p[1];
  for (int i = 2; i <= tn; i++) {
    int u = p[i];
    for (; top > 1 && out[u] > out[stk[top]]; top--)
      adl(stk[top - 1], stk[top], dep[stk[top]] - dep[stk[top - 1]]);
    stk[++top] = u;
  }
  for (; top > 1; top--)
    adl(stk[top - 1], stk[top], dep[stk[top]] - dep[stk[top - 1]]);
  for (int i = 1; i <= tn; i++) dis[p[i]] = INF;
  for (int i = 1; i <= m0; i++) {
    dis[p0[i]] = dis2[p0[i]] = 0;
    col[p0[i]] = i;
    q.push((Data){0, i, p0[i]});
  }
  while (!q.empty()) {
    Data now = q.top();
    q.pop();
    if (dis[now.p] != now.d || col[now.p] != now.c) continue;
    int u = now.p, c = now.c;
    for (int i = fir[u]; i; i = l[i].nxt) {
      int d = (dis2[u] + l[i].l - 1) / w[c] + 1, v = l[i].t;
      if (dis[v] > d || (dis[v] == d && col[v] > c)) {
        dis2[v] = dis2[u] + l[i].l;
        q.push((Data){dis[v] = d, col[v] = c, v});
      }
    }
  }
  for (int i = 1; i <= m1; i++) printf("%d ", col[p1[i]]);
  puts("");
  tl = 0;
  for (int i = 1; i <= tn; i++) fir[p[i]] = 0;
}
int n, m;
int main() {
  n = read();
  for (int i = 1, u, v; i < n; i++) {
    u = read();
    v = read();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dep[1] = 1;
  pfs1(1);
  pfs2(1, 1);
  m = read();
  while (m--) solve();
  return 0;
}