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 = 2e5 + 9;
int n, q, k, m;
int d[N], s[N], w[N];
namespace frame {
int head[N], tot;
struct pp {
  int nxt, to, val;
} g[N << 1];
void add(int u, int v, int w) {
  g[++tot].nxt = head[u], g[tot].to = v, g[tot].val = w, head[u] = tot;
  return;
}
void E(int u, int v, int w) {
  if (w < 0) throw;
  add(u, v, w);
  add(v, u, w);
  return;
}
struct S {
  int a, t, k, res;
  S(int aa = 0, int tt = 0, int kk = 0, int rr = 0) {
    a = aa, t = tt, k = kk, res = rr;
  }
} p[N];
bool operator<(S a, S b) { return (a.t == b.t) ? (a.k > b.k) : (a.t > b.t); }
priority_queue<S> q;
int c[N];
void framework() {
  memset(head, -1, sizeof(head));
  tot = 0;
  memset(c, 0, sizeof(c));
  for (int i = 1; i <= n; i++) p[i].t = -1;
  return;
}
void solve() {
  for (int i = 1; i <= k; i++) {
    q.push(S(d[i], 0, i, 0));
    p[d[i]] = S(0, 0, i, 0);
    c[d[i]] = i;
  }
  while (!q.empty()) {
    S u = q.top();
    q.pop();
    if (c[u.a] && u.t) continue;
    c[u.a] = u.k;
    for (int i = head[u.a]; i != -1; i = g[i].nxt) {
      int v = g[i].to, tt = g[i].val - u.res;
      if (tt < 0)
        tt = u.t;
      else {
        tt = tt / s[u.k] + (tt % s[u.k] > 0) + u.t;
      }
      if (p[v].t == -1 || p[v].t > tt || (p[v].t == tt && p[v].k > p[u.a].k)) {
        if (c[v]) throw;
        p[v].t = tt, p[v].k = p[u.a].k;
        q.push(S(v, tt, p[v].k, (tt - u.t) * s[u.k] + u.res - g[i].val));
      }
    }
  }
  for (int i = 1; i <= m; i++) printf("%d ", c[w[i]]);
  puts("");
  return;
}
}  // namespace frame
namespace te {
int head[N], tot;
struct pp {
  int nxt, to;
} g[N << 1];
void add(int u, int v) { g[++tot].nxt = head[u], g[tot].to = v, head[u] = tot; }
int dfn[N], dep[N], son[N], sz[N], pa[N], num;
void dfs1(int u, int fa) {
  sz[u] = 1, dep[u] = dep[fa] + 1, dfn[u] = ++num, pa[u] = fa;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
  }
  return;
}
int top[N];
void dfs2(int u, int fa, int topf) {
  top[u] = topf;
  if (son[u]) dfs2(son[u], u, topf);
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    ;
    if (v == fa || v == son[u]) continue;
    dfs2(v, u, v);
  }
  return;
}
int lca(int u, int v) {
  while (top[u] != top[v]) {
    if (dep[top[u]] < dep[top[v]]) swap(u, v);
    u = pa[top[u]];
  }
  return dep[u] > dep[v] ? v : u;
}
void bl() {
  memset(head, -1, sizeof(head));
  tot = 0;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  m = 0;
  dfs1(1, 0);
  dfs2(1, 0, 1);
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
int s[N], t, w[N], m, c[N], l;
void foundation() {
  m = ::m;
  for (int i = 1; i <= m; i++) w[i] = ::w[i];
  for (int i = 1; i <= k; i++) w[++m] = d[i];
  sort(w + 1, w + m + 1, cmp);
  t = 0;
  m = unique(w + 1, w + m + 1) - w - 1;
  l = m;
  for (int i = 1; i <= m; i++) c[i] = w[i];
  for (int i = 1; i <= m; i++) {
    if (t) {
      int u = lca(s[t], w[i]), former = 0;
      while (t && dep[s[t]] > dep[u]) {
        if (former) frame::E(s[t], former, dep[former] - dep[s[t]]);
        former = s[t];
        s[t--] = 0;
      }
      if (s[t] != u) s[++t] = u, c[++l] = u;
      if (former) frame::E(u, former, dep[former] - dep[u]);
    }
    s[++t] = w[i];
  }
  while (t) {
    if (t > 1) frame::E(s[t - 1], s[t], dep[s[t]] - dep[s[t - 1]]);
    s[t--] = 0;
  }
  return;
}
void clear() {
  for (int i = 1; i <= l; i++) {
    frame::head[c[i]] = -1;
    frame::c[c[i]] = 0, frame::p[c[i]] = frame::S(0, -1, 0, 0);
  }
  return;
}
}  // namespace te
int main() {
  te::bl();
  scanf("%d", &q);
  frame::framework();
  while (q--) {
    scanf("%d%d", &k, &m);
    for (int i = 1; i <= k; i++) scanf("%d%d", &d[i], &s[i]);
    for (int i = 1; i <= m; i++) scanf("%d", &w[i]);
    te::tot = 0;
    frame::tot = 0;
    te::foundation();
    frame::solve();
    te::clear();
  }
  return 0;
}