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;
int n, q, cnt, a[N], b[N], f[N], p[N], s[N], fa[N], dep[N], siz[N], dfn[N],
    top[N], stk[N], c[N * 4];
vector<int> e[N], E[N];
void dfs1(int u) {
  siz[u] = 1;
  for (int v : e[u])
    if (v != fa[u]) fa[v] = u, dep[v] = dep[u] + 1, dfs1(v), siz[u] += siz[v];
}
void dfs2(int u) {
  dfn[u] = ++cnt;
  int w = 0;
  for (int v : e[u])
    if (v != fa[u] && siz[v] > siz[w]) w = v;
  if (w) top[w] = top[u], dfs2(w);
  for (int v : e[u])
    if (v != fa[u] && v != w) top[v] = v, dfs2(v);
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
int lca(int x, int y) {
  while (top[x] != top[y])
    if (dep[top[x]] > dep[top[y]])
      x = fa[top[x]];
    else
      y = fa[top[y]];
  return dep[x] < dep[y] ? x : y;
}
bool calc(int a, int b, int c, int d) {
  int x = (c + s[d] - 1) / s[d] - (a + s[b] - 1) / s[b];
  return x > 0 || (!x && b < d);
}
void dfs3(int u) {
  for (int v : E[u]) {
    dfs3(v);
    if (p[v] && (!p[u] || calc(f[v] + dep[v] - dep[u], p[v], f[u], p[u])))
      p[u] = p[v], f[u] = f[v] + dep[v] - dep[u];
  }
}
void dfs4(int u) {
  for (int v : E[u]) {
    if (p[u] && (!p[v] || calc(f[u] + dep[v] - dep[u], p[u], f[v], p[v])))
      p[v] = p[u], f[v] = f[u] + dep[v] - dep[u];
    dfs4(v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; ++i)
    scanf("%d%d", &x, &y), e[x].push_back(y), e[y].push_back(x);
  dfs1(1), top[1] = 1, dfs2(1);
  scanf("%d", &q);
  while (q--) {
    int k, m, x, y;
    scanf("%d%d", &k, &m);
    for (int i = 1; i <= k; ++i)
      scanf("%d%d", a + i, s + i), c[i] = a[i], p[a[i]] = i;
    for (int i = 1; i <= m; ++i) scanf("%d", b + i), c[i + k] = b[i];
    y = x = k + m, sort(c + 1, c + x + 1, cmp);
    int t = 1;
    stk[1] = 1;
    for (int i = 1; i <= x; ++i) {
      int u = lca(stk[t], c[i]);
      c[++y] = u;
      while (t > 1 && dep[stk[t - 1]] >= dep[u])
        E[stk[t - 1]].push_back(stk[t]), --t;
      if (stk[t] == u) --t;
      if (dep[stk[t]] > dep[u]) E[u].push_back(stk[t]), --t;
      stk[++t] = u;
      if (u != c[i]) stk[++t] = c[i];
    }
    while (t > 1) E[stk[t - 1]].push_back(stk[t]), --t;
    dfs3(1), dfs4(1);
    for (int i = 1; i <= m; ++i) printf("%d%c", p[b[i]], " \n"[i == m]);
    for (int i = 1; i <= y; ++i) p[c[i]] = f[c[i]] = 0, E[c[i]].clear();
  }
  return 0;
}