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 = 200005, K = 20;
int n, Q, k, m, ind, top, tin[N], v[N], s[N], u[N], num[N * 2], dep[N], f[N][K],
    st[N], ans[N], tim[N], ord[N];
struct ed {
  int nxt, to;
} G[N << 1];
int lnk[N], pp;
void ae(int k1, int k2) { G[++pp] = (ed){lnk[k1], k2}, lnk[k1] = pp; }
void init(int k1, int k2) {
  int i, j;
  tin[k1] = ++ind;
  for (i = 1; i < K; ++i) f[k1][i] = f[f[k1][i - 1]][i - 1];
  for (i = lnk[k1]; i; i = G[i].nxt)
    if ((j = G[i].to) != k2) f[j][0] = k1, dep[j] = dep[k1] + 1, init(j, k1);
}
int lca(int k1, int k2) {
  int i;
  if (dep[k1] < dep[k2]) swap(k1, k2);
  int ret = dep[k1] - dep[k2];
  for (i = K - 1; i >= 0; --i)
    if (ret >> i & 1) k1 = f[k1][i];
  if (k1 == k2) return k1;
  for (i = K - 1; i >= 0; --i)
    if (f[k1][i] != f[k2][i]) k1 = f[k1][i], k2 = f[k2][i];
  return f[k1][0];
}
int dist(int k1, int k2) { return dep[k1] + dep[k2] - dep[lca(k1, k2)] * 2; }
bool cmp(int k1, int k2) { return tin[k1] < tin[k2]; }
void up(int k1, int k2) {
  if (k1) {
    int len = (dist(k1, k2) + s[k1] - 1) / s[k1];
    if (len < tim[k2] || (len == tim[k2] && ord[k1] < ord[ans[k2]]))
      ans[k2] = k1, tim[k2] = len;
  }
}
void dfs(int k1) {
  int i, j;
  for (i = lnk[k1]; i; i = G[i].nxt) {
    j = G[i].to;
    up(ans[k1], j);
    dfs(j);
  }
}
void dfs2(int k1) {
  int i, j;
  for (i = lnk[k1]; i; i = G[i].nxt) {
    j = G[i].to;
    dfs2(j);
    up(ans[j], k1);
  }
}
void del(int k1) {
  int i, j;
  for (i = lnk[k1]; i; i = G[i].nxt) {
    j = G[i].to;
    del(j);
  }
  lnk[k1] = 0, ans[k1] = 0, tim[k1] = 0x3f3f3f3f, ord[k1] = 0;
}
int main() {
  int i, k1, k2, tot;
  memset(tim, 63, sizeof(tim));
  scanf("%d", &n);
  for (i = 1; i < n; ++i) scanf("%d%d", &k1, &k2), ae(k1, k2), ae(k2, k1);
  init(1, 0);
  memset(lnk, 0, sizeof(lnk)), pp = 0;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &k, &m);
    tot = top = 0;
    for (i = 1; i <= k; ++i)
      scanf("%d", &v[i]), scanf("%d", &s[v[i]]),
          num[++tot] = v[i], ans[v[i]] = v[i], tim[v[i]] = 0, ord[v[i]] = i;
    for (i = 1; i <= m; ++i) scanf("%d", &u[i]), num[++tot] = u[i];
    sort(num + 1, num + 1 + tot, cmp);
    st[++top] = 1;
    for (i = 1; i <= tot; ++i) {
      int l = lca(st[top], num[i]);
      while (top > 1 && dep[st[top - 1]] >= dep[l])
        ae(st[top - 1], st[top]), --top;
      if (st[top] != l) ae(l, st[top]);
      st[top] = l;
      if (l != num[i]) st[++top] = num[i];
    }
    for (i = 1; i < top; ++i) ae(st[i], st[i + 1]);
    dfs2(1);
    dfs(1);
    for (i = 1; i <= m; ++i)
      printf("%d%c", ord[ans[u[i]]], i == m ? '\n' : ' ');
    del(1);
    pp = 0;
  }
  return 0;
}