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
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkmax(T &a, const T &b) {
  a = a > b ? a : b;
}
template <typename T>
void chkmin(T &a, const T &b) {
  a = a < b ? a : b;
}
const int MAXN = 200005, MAXM = 400005;
vector<int> gra[MAXN], tr[MAXN];
int arr[MAXM], st[20][MAXM], lg[MAXM], dfn[MAXN], typ[MAXN];
int dep[MAXN], spd[MAXN], qry[MAXN], sta[MAXN], fuck[MAXN], n, Q, tot;
void dfs1(int u, int fa) {
  st[0][dfn[u] = ++tot] = u;
  for (int v : gra[u]) {
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    dfs1(v, u);
    st[0][++tot] = u;
  }
}
int get_min(int x, int y) { return dep[x] < dep[y] ? x : y; }
int get_lca(int x, int y) {
  x = dfn[x], y = dfn[y];
  if (x > y) swap(x, y);
  int l = lg[y - x + 1];
  return get_min(st[l][x], st[l][y - (1 << l) + 1]);
}
int get_dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[get_lca(x, y)]; }
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
struct P {
  int t, d, p;
} f[MAXN];
void gmin(P &a, const P &b, int d) {
  if (b.p > MAXN) return;
  int t = (b.d + d + spd[b.p] - 1) / spd[b.p];
  if (t < a.t || (t == a.t && b.p < a.p)) a.d = b.d + d, a.t = t, a.p = b.p;
}
void dfs2(int u) {
  f[u] = typ[u] ? P{0, 0, typ[u]} : P{(int)1e9, (int)1e9, (int)1e9};
  for (int v : tr[u]) {
    dfs2(v);
    gmin(f[u], f[v], dep[v] - dep[u]);
  }
}
void dfs3(int u) {
  if (f[u].p < MAXN) assert(f[u].d >= get_dis(u, fuck[f[u].p]));
  for (int v : tr[u]) {
    gmin(f[v], f[u], dep[v] - dep[u]);
    dfs3(v);
  }
  tr[u].clear();
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    gra[u].push_back(v);
    gra[v].push_back(u);
  }
  dfs1(1, 0);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j + (1 << i) - 1 <= tot; j++)
      st[i][j] = get_min(st[i - 1][j], st[i - 1][j + (1 << i >> 1)]);
  for (int i = 2; i <= tot; i++) lg[i] = lg[i >> 1] + 1;
  for (scanf("%d", &Q); Q--;) {
    int ki, mi;
    scanf("%d%d", &ki, &mi);
    tot = 0;
    for (int i = 1; i <= ki; i++) {
      int v, s;
      scanf("%d%d", &v, &s);
      spd[i] = s;
      typ[v] = i;
      fuck[i] = v;
      arr[++tot] = v;
    }
    for (int i = 1; i <= mi; i++) {
      scanf("%d", qry + i);
      arr[++tot] = qry[i];
    }
    arr[++tot] = 1;
    sort(arr + 1, arr + 1 + tot, cmp);
    tot = unique(arr + 1, arr + 1 + tot) - arr - 1;
    int tp = 0;
    sta[++tp] = 1;
    for (int i = 2; i <= tot; i++) {
      int l = get_lca(arr[i], sta[tp]);
      for (; tp > 1 && dep[l] <= dep[sta[tp - 1]]; --tp)
        tr[sta[tp - 1]].push_back(sta[tp]);
      if (l != sta[tp]) {
        tr[l].push_back(sta[tp]);
        sta[tp] = l;
      }
      sta[++tp] = arr[i];
    }
    for (; tp > 1; --tp) tr[sta[tp - 1]].push_back(sta[tp]);
    dfs2(1);
    dfs3(1);
    for (int i = 1; i <= mi; i++) {
      int g = fuck[f[qry[i]].p];
      assert(f[qry[i]].d == dep[qry[i]] + dep[g] - 2 * dep[get_lca(qry[i], g)]);
      printf("%d ", f[qry[i]].p);
    }
    puts("");
    for (int i = 1; i <= tot; i++) typ[arr[i]] = 0;
  }
  return 0;
}
