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 SIZE = 2e5 + 5;
const int inf = 1e9;
int n, q, ki, mi, num, tim, cnt, top;
int head[SIZE], dfn[SIZE], siz[SIZE], son[SIZE], bel[SIZE], fa[SIZE], dep[SIZE];
int sta[SIZE], s[SIZE], v[SIZE], key[SIZE], vis[SIZE];
std::vector<int> vir;
struct tree {
  int to, w;
};
std::vector<tree> g[SIZE];
struct node {
  int to, nxt;
} edge[SIZE << 1];
struct ele {
  int tim, dis, col, idx;
  bool operator<(const ele &a) const {
    return tim ^ a.tim ? tim > a.tim : col > a.col;
  }
} dis[SIZE];
std::priority_queue<ele> que;
namespace GTR {
const int bufl = 1 << 15;
char buf[bufl], *s = buf, *t = buf;
inline int fetch() {
  if (s == t) {
    t = (s = buf) + fread(buf, 1, bufl, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
inline int read() {
  int a = 0, b = 1, c = fetch();
  while (c < 48 || c > 57) b ^= c == '-', c = fetch();
  while (c >= 48 && c <= 57) a = (a << 1) + (a << 3) + c - 48, c = fetch();
  return b ? a : -a;
}
}  // namespace GTR
using GTR::read;
void addEdge(int u, int v) { edge[++num] = (node){v, head[u]}, head[u] = num; }
void build(int u, int v, int d) {
  g[u].emplace_back((tree){v, d}), g[v].emplace_back((tree){u, d});
}
void dfs1(int u, int anc) {
  siz[u] = 1;
  for (int i = head[u], v; i; i = edge[i].nxt) {
    v = edge[i].to;
    if (v == anc) continue;
    fa[v] = u, dep[v] = dep[u] + 1;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void dfs2(int u, int top) {
  dfn[u] = ++tim, bel[u] = top;
  if (son[u]) dfs2(son[u], top);
  for (int i = head[u], v; i; i = edge[i].nxt) {
    v = edge[i].to;
    if (v == son[u] || v == fa[u]) continue;
    dfs2(v, v);
  }
}
int lca(int u, int v) {
  while (bel[u] ^ bel[v]) {
    if (dep[bel[u]] > dep[bel[v]]) std::swap(u, v);
    v = fa[bel[v]];
  }
  return dep[u] < dep[v] ? u : v;
}
int cmp(int a, int b) { return dfn[a] < dfn[b]; }
void makeVir() {
  std::sort(vir.begin(), vir.end(), cmp);
  for (int i = vir.size() - 2; ~i; --i)
    vir.emplace_back(lca(vir[i], vir[i + 1]));
  std::sort(vir.begin(), vir.end(), cmp);
  vir.erase(unique(vir.begin(), vir.end()), vir.end()), top = 0;
  for (int u : vir) {
    while (top && dfn[sta[top]] + siz[sta[top]] - 1 < dfn[u]) --top;
    if (top) build(u, sta[top], abs(dep[u] - dep[sta[top]]));
    sta[++top] = u;
  }
  for (int u : vir) dis[u] = (ele){inf, inf, 0, u};
}
void dijkstra() {
  for (int i = 1; i <= ki; ++i) que.push(dis[s[i]] = (ele){0, 0, i, s[i]});
  while (!que.empty()) {
    ele head = que.top();
    que.pop();
    int u = head.idx, disU = head.dis, colU = head.col;
    if (vis[u]) continue;
    vis[u] = 1;
    for (auto it : g[u]) {
      int ndi = disU + it.w;
      ele tail;
      tail = (ele){(ndi + v[colU] - 1) / v[colU], ndi, colU, it.to};
      if (dis[it.to] < tail) dis[it.to] = tail, que.push(tail);
    }
  }
}
int main() {
  n = read();
  for (int i = 1, u, v; i < n; ++i) {
    u = read(), v = read();
    addEdge(u, v), addEdge(v, u);
  }
  dfs1(1, 0), dfs2(1, 1);
  q = read();
  while (q--) {
    ki = read(), mi = read();
    for (int i = 1; i <= ki; ++i)
      s[i] = read(), v[i] = read(), vir.emplace_back(s[i]);
    for (int i = 1; i <= mi; ++i) key[i] = read(), vir.emplace_back(key[i]);
    makeVir();
    dijkstra();
    for (int i = 1; i <= mi; ++i) printf("%d ", dis[key[i]].col);
    puts("");
    for (int u : vir) g[u].clear(), vis[u] = 0;
    vir.clear();
  }
  return 0;
}