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 ll = int;
using ld = long double;
using ull = unsigned long long;
const ll mod = 1e9 + 7;
using namespace std;
const int N = 2e5 + 5;
ll m, n, k, tong, t, T, ans, u, v, q;
ll b[N], f[N], pos[N], d[N], in[N], out[N], par[N], chain, P[N][19], h[N];
string s;
pair<ll, ll> a[N], res[N];
vector<ll> adj[N];
vector<pair<ll, ll> > kq, adt[N];
void dfs(ll u) {
  for (int i = 1; i <= 18; i++) P[u][i] = P[P[u][i - 1]][i - 1];
  in[u] = ++m;
  for (ll v : adj[u]) {
    if (v != par[u]) {
      h[v] = h[u] + 1;
      P[v][0] = u;
      par[v] = u;
      dfs(v);
      d[u] += d[v];
    }
  }
  out[u] = m;
}
bool ins(ll x, ll y) { return (in[y] <= in[x] && out[x] <= out[y]); }
ll lca(ll u, ll v) {
  if (h[u] < h[v]) swap(u, v);
  ll log = log2(h[u] + 1);
  for (int i = log; i >= 0; i--)
    if (h[u] >= h[v] + (1ll << i)) u = P[u][i];
  if (u == v) return u;
  for (int i = log; i >= 0; i--) {
    if (P[u][i] != 0 && P[u][i] != P[v][i]) {
      u = P[u][i];
      v = P[v][i];
    }
  }
  return par[u];
}
ll dis(ll u, ll v) {
  ll z = lca(u, v);
  return h[u] + h[v] - 2 * h[z];
}
ll tim(ll dis, ll v) { return (dis + v - 1) / v; }
void query() {
  priority_queue<pair<pair<ll, ll>, ll>, vector<pair<pair<ll, ll>, ll> >,
                 greater<pair<pair<ll, ll>, ll> > >
      pq;
  cin >> k >> m;
  kq.clear();
  for (int i = 1; i <= k; i++) {
    cin >> a[i].first >> a[i].second;
    pair<pair<ll, ll>, ll> u;
    u.first.first = 0;
    u.first.second = i;
    u.second = a[i].first;
    res[a[i].first].first = 0;
    res[a[i].first].second = i;
    pq.push(u);
    kq.push_back({in[a[i].first], a[i].first});
  }
  for (int i = 1; i <= m; i++) {
    cin >> b[i];
    kq.push_back({in[b[i]], b[i]});
  }
  sort(kq.begin(), kq.end());
  kq.erase(unique(kq.begin(), kq.end()), kq.end());
  t = kq.size();
  for (int i = 0; i < t; i++) {
    pair<ll, ll> u = kq[i];
    if (i) {
      pair<ll, ll> v = kq[i - 1];
      ll z = lca(u.second, v.second);
      kq.push_back({in[z], z});
    }
  }
  sort(kq.begin(), kq.end());
  kq.erase(unique(kq.begin(), kq.end()), kq.end());
  stack<ll> ds;
  ds.push(kq[0].second);
  for (int i = 1; i < kq.size(); i++) {
    while (!ds.empty()) {
      if (ins(kq[i].second, ds.top())) {
        adt[ds.top()].push_back({dis(ds.top(), kq[i].second), kq[i].second});
        adt[kq[i].second].push_back({dis(ds.top(), kq[i].second), ds.top()});
        break;
      } else
        ds.pop();
    }
    ds.push(kq[i].second);
  }
  while (!pq.empty()) {
    pair<pair<ll, ll>, ll> u = pq.top();
    pq.pop();
    if (res[u.second].first != u.first.first ||
        res[u.second].second != u.first.second)
      continue;
    for (pair<ll, ll> v : adt[u.second]) {
      if ((res[v.second].first >
           tim(dis(a[res[u.second].second].first, v.second),
               a[res[u.second].second].second)) ||
          ((res[v.second].first ==
                tim(dis(a[res[u.second].second].first, v.second),
                    a[res[u.second].second].second) &&
            res[v.second].second > res[u.second].second))) {
        res[v.second].first = tim(dis(a[res[u.second].second].first, v.second),
                                  a[res[u.second].second].second);
        res[v.second].second = res[u.second].second;
        pair<pair<ll, ll>, ll> x;
        x.first.first = res[v.second].first;
        x.first.second = res[v.second].second;
        x.second = v.second;
        pq.push(x);
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << res[b[i]].second << " ";
  }
  cout << '\n';
  for (int i = 0; i < kq.size(); i++) {
    adt[kq[i].second].clear();
    res[kq[i].second].first = mod;
    res[kq[i].second].second = mod;
  }
}
void sol() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    res[i].first = mod;
    res[i].second = mod;
  }
  for (int i = 1; i < n; i++) {
    ll x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  h[1] = 1;
  dfs(1);
  cin >> q;
  while (q-- > 0) {
    query();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("cowland"
            ".in",
            "r")) {
    freopen(
        "cowland"
        ".in",
        "r", stdin);
    freopen(
        "cowland"
        ".out",
        "w", stdout);
  }
  sol();
}