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 + 10;
const int LOG = 21;
int n, tim, t;
vector<int> E[N], g[N];
int dep[N], f[N][LOG], dfn[N], s[N];
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
inline void dfs(int x) {
  dfn[x] = ++tim;
  for (auto y : E[x]) {
    if (y != f[x][0]) {
      dep[y] = dep[x] + 1;
      f[y][0] = x;
      for (int i = 1; f[y][i - 1]; ++i) f[y][i] = f[f[y][i - 1]][i - 1];
      dfs(y);
    }
  }
}
inline int lca(int x, int y) {
  if (dep[x] > dep[y]) swap(x, y);
  for (int i = LOG - 1; i >= 0; --i)
    if (dep[f[y][i]] >= dep[x]) y = f[y][i];
  if (x == y) return x;
  for (int i = LOG - 1; i >= 0; --i)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
inline void Virtual_Tree(vector<int> &ptr) {
  sort(ptr.begin(), ptr.end(), [&](int x, int y) { return dfn[x] < dfn[y]; });
  ptr.erase(unique(ptr.begin(), ptr.end()), ptr.end());
  vector<int> q = ptr;
  for (auto x : ptr) {
    int lca = ::lca(x, s[t]);
    if (lca != s[t]) {
      while (t && dep[s[t - 1]] >= dep[lca])
        g[s[t]].push_back(s[t - 1]), g[s[t - 1]].push_back(s[t]), --t;
      if (s[t] != lca)
        g[s[t]].push_back(lca), g[lca].push_back(s[t]), s[t] = lca,
                                                        q.push_back(lca);
    }
    s[++t] = x;
  }
  if (t)
    while (--t) g[s[t]].push_back(s[t + 1]), g[s[t + 1]].push_back(s[t]);
  ptr = q;
}
int w[N];
bool v[N];
struct P {
  int d, i, x, r;
  inline P() {}
  inline P(int d, int i, int x) : d(d), i(i), x(x) {
    r = d ? (d - 1) / w[x] : -1;
  }
  inline bool operator<(const P o) const {
    return r == o.r ? i < o.i : r < o.r;
  }
  inline P operator-() const {
    P o = *this;
    o.r *= -1, o.i *= -1;
    return o;
  }
} u[N];
inline void dijkstra(vector<int> a, vector<int> b, vector<int> p) {
  priority_queue<pair<P, int> > q;
  for (int x : p) u[x].r = u[x].i = n, v[x] = 0;
  for (unsigned i = 0; i < a.size(); i++)
    u[a[i]] = P(0, i, a[i]), q.push(make_pair(-u[a[i]], a[i]));
  while (!q.empty()) {
    int x = q.top().second;
    q.pop();
    if (v[x]) continue;
    v[x] = 1;
    for (int y : g[x]) {
      P o = P(u[x].d + abs(dep[x] - dep[y]), u[x].i, u[x].x);
      if (o < u[y]) u[y] = o, q.push(make_pair(-u[y], y));
    }
  }
  for (int x : b) printf("%d ", u[x].i + 1);
  printf("\n");
}
int main() {
  read(n);
  for (int i = 1, x, y; i < n; ++i) {
    read(x), read(y);
    E[x].push_back(y);
    E[y].push_back(x);
  }
  dep[1] = 1;
  dfs(1);
  int q;
  read(q);
  while (q--) {
    int A, B;
    read(A);
    read(B);
    vector<int> a, b, p;
    for (int i = 1, x, y; i <= A; ++i)
      read(x), read(y), a.push_back(x), p.push_back(x), w[x] = y;
    for (int i = 1, x; i <= B; ++i) read(x), b.push_back(x), p.push_back(x);
    Virtual_Tree(p);
    dijkstra(a, b, p);
    for (auto x : p) g[x].clear();
  }
  return 0;
}