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 = 1000000 + 5;
struct edge {
  int to, val, nxt;
} e[N << 1];
vector<int> v[N];
int dep[N], son[N], fa[N], head[N], cnt, sz[N], top[N], tot, dfn[N];
void add(int x, int y, int z) {
  e[++cnt].to = y;
  e[cnt].val = z;
  e[cnt].nxt = head[x];
  head[x] = cnt;
}
void dfs2(int x, int topfa) {
  dfn[x] = ++tot;
  top[x] = topfa;
  if (son[x]) dfs2(son[x], topfa);
  for (int i = head[x]; i; i = e[i].nxt)
    if (!dfn[e[i].to]) {
      dfs2(e[i].to, e[i].to);
    }
}
void dfs1(int x) {
  sz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt)
    if (e[i].to != fa[x]) {
      fa[e[i].to] = x;
      dep[e[i].to] = dep[x] + 1;
      dfs1(e[i].to);
      sz[x] += sz[e[i].to];
      if (sz[e[i].to] > sz[son[x]]) son[x] = e[i].to;
    }
}
int LCA(int x, int y) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  return x;
}
void add_edge(int x, int y) { v[x].push_back(y); }
int rd() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f = (ch == '-' ? -1 : f), ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int key[N], s[N], Top, K, a[N], b[N], ans[N];
int col[N], f[N], pos[N], spd[N];
void insert(int x) {
  if (x == 1) return;
  int lca = LCA(x, s[Top]);
  while (Top > 1 && dfn[s[Top - 1]] >= dfn[lca]) {
    add_edge(s[Top - 1], s[Top]), --Top;
  }
  if (lca ^ s[Top]) add_edge(lca, s[Top--]), s[++Top] = lca;
  s[++Top] = x;
}
bool cmp(int i, int j) { return dfn[i] < dfn[j]; }
void solve() {
  sort(a + 1, a + K + 1, cmp);
  s[Top = 1] = 1;
  for (int i = 1; i <= K; ++i)
    if (a[i] != a[i - 1]) insert(a[i]);
  for (int i = 1; i < Top; ++i) add_edge(s[i], s[i + 1]);
}
inline int dis(int x, int y) {
  int lca = LCA(x, y);
  return dep[x] + dep[y] - (dep[lca] << 1);
}
void redfs(int x) {
  int tp1, tp2;
  for (int to : v[x]) {
    if (col[x]) {
      if (!col[to])
        col[to] = col[x];
      else {
        tp1 = ceil(1.0 * dis(to, pos[col[x]]) / spd[col[x]]),
        tp2 = ceil(1.0 * dis(to, pos[col[to]]) / spd[col[to]]);
        if (tp1 < tp2 || (tp1 == tp2 && col[x] < col[to])) col[to] = col[x];
      }
    }
    redfs(to);
  }
  v[x].clear();
  ans[x] = col[x];
  col[x] = 0;
}
void dfs(int x) {
  int tp1, tp2;
  for (int to : v[x]) {
    dfs(to);
    if (col[to]) {
      if (!col[x])
        col[x] = col[to];
      else {
        tp1 = ceil(1.0 * dis(x, pos[col[x]]) / spd[col[x]]),
        tp2 = ceil(1.0 * dis(x, pos[col[to]]) / spd[col[to]]);
        if (tp1 > tp2 || (tp1 == tp2 && col[x] > col[to])) col[x] = col[to];
      }
    }
  }
}
int main() {
  int x, y, n = rd();
  for (int i = 1; i < n; ++i) {
    x = rd();
    y = rd();
    add(x, y, 1);
    add(y, x, 1);
  }
  dep[1] = 1;
  dfs1(1);
  dfs2(1, 1);
  int q = rd();
  int m, k;
  while (q--) {
    k = rd();
    m = rd();
    K = 0;
    for (int i = 1; i <= k; ++i)
      x = rd(), y = rd(), a[++K] = x, pos[i] = x, spd[i] = y, col[x] = i;
    for (int i = 1; i <= m; ++i) b[i] = rd(), a[++K] = b[i];
    solve();
    dfs(1);
    redfs(1);
    for (int i = 1; i <= m; ++i) cout << ans[b[i]] << " ";
    cout << endl;
  }
  return 0;
}