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 maxn = 1e6 + 999;
vector<int> v[maxn];
vector<pair<int, int> > g[maxn];
int fa[maxn], dep[maxn], siz[maxn], Top[maxn], son[maxn], dfn[maxn], clk;
void dfs1(int x) {
  siz[x] = 1;
  dfn[x] = ++clk;
  for (int y : v[x]) {
    if (y == fa[x]) continue;
    fa[y] = x;
    dep[y] = dep[x] + 1;
    dfs1(y);
    siz[x] += siz[y];
    if (siz[y] > siz[son[x]]) son[x] = y;
  }
}
void dfs2(int x, int topf) {
  Top[x] = topf;
  if (son[x]) dfs2(son[x], topf);
  for (int y : v[x]) {
    if (y != fa[x] && y != son[x]) dfs2(y, y);
  }
}
int LCA(int x, int y) {
  while (Top[x] ^ Top[y])
    if (dep[Top[x]] < dep[Top[y]])
      y = fa[Top[y]];
    else
      x = fa[Top[x]];
  return dep[x] > dep[y] ? y : x;
}
int n, st[maxn], V[maxn], imp[maxn], top, a[maxn], s[maxn];
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
inline int dis(int x, int y) {
  int lca = LCA(x, y);
  return dep[x] + dep[y] - 2 * dep[lca];
}
void add(int x, int y) {
  int t = dis(x, y);
  g[x].push_back(make_pair(y, t));
  g[y].push_back(make_pair(x, t));
}
void insert(int x) {
  int lca = LCA(x, s[top]);
  while (top > 1 && dfn[s[top - 1]] >= dfn[lca]) add(s[top - 1], s[top]), --top;
  if (lca != s[top]) add(lca, s[top]), s[top] = lca;
  s[++top] = x;
}
struct node {
  int x, t, id, d, fl;
  friend bool operator<(node a, node b) {
    return a.t == b.t ? (a.id == b.id ? a.d > b.d : a.id < b.id) : a.t < b.t;
  }
} Dis[maxn];
struct nd {
  node first;
  int second;
  friend bool operator<(nd x, nd y) { return y.first < x.first; }
};
priority_queue<nd> q;
int vis[maxn];
vector<int> tmp;
void dfs3(int x, int faa) {
  tmp.push_back(x);
  for (auto now : g[x]) {
    int y = now.first;
    if (y == faa) continue;
    dfs3(y, x);
  }
}
int n1, n2;
node To(int x, pair<int, int> y) {
  node ret = Dis[x];
  if (y.second <= ret.d) {
    ret.d -= y.second;
    return ret;
  }
  int dist = y.second - ret.d;
  ret.t += (dist + V[ret.id] - 1) / V[ret.id];
  ret.d = (V[ret.id] * ((dist + V[ret.id] - 1) / V[ret.id]) - dist);
  return ret;
}
void dj() {
  while (q.size()) q.pop();
  for (int i = (1); i <= (n1); ++i)
    Dis[st[i]] = node{st[i], 0, i, 0, 1}, q.push(nd{Dis[st[i]], st[i]});
  while (q.size()) {
    int x = q.top().second;
    q.pop();
    if (vis[x]) continue;
    vis[x] = 1;
    for (auto y : g[x]) {
      node t = To(x, y);
      if ((!Dis[y.first].fl) || t < Dis[y.first])
        Dis[y.first] = t, q.push(nd{Dis[y.first], y.first});
    }
  }
  for (int i = (1); i <= (n2); ++i) cout << Dis[imp[i]].id << " ";
  cout << endl;
}
void init() {
  top = 0;
  for (int x : tmp) g[x].clear(), vis[x] = 0, Dis[x] = node{0, 0, 0, 0};
  tmp.clear();
}
int main() {
  cin >> n;
  for (int i = (1); i <= (n - 1); ++i) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs1(1);
  dfs2(1, 1);
  int q;
  cin >> q;
  while (q--) {
    cin >> n1 >> n2;
    for (int i = (1); i <= (n1); ++i) cin >> st[i] >> V[i], a[i] = st[i];
    for (int i = (1); i <= (n2); ++i) cin >> imp[i], a[i + n1] = imp[i];
    sort(a + 1, a + 1 + n1 + n2);
    int len = unique(a + 1, a + 1 + n1 + n2) - a - 1;
    sort(a + 1, a + 1 + len, cmp);
    s[top = 1] = 1;
    for (int i = (1); i <= (len); ++i)
      if (a[i] != 1) insert(a[i]);
    while (top > 1) add(s[top - 1], s[top]), --top;
    dj();
    dfs3(1, 0);
    init();
  }
}