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;
struct node {
  int t, next;
} a[400010];
vector<int> v[200010];
pair<int, int> h[200010];
int num[200010];
int fa[200010][20], dep[200010];
int head[200010], dfn[200010], c[400010], n, m, cl, cnt, tot;
inline int rd() {
  int x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x;
}
inline void print(int x) {
  static char s[233];
  if (!x) {
    putchar('0');
    putchar(' ');
    return;
  }
  int tot = 0;
  for (; x; x /= 10) s[++tot] = x % 10 + '0';
  for (; tot; tot--) putchar(s[tot]);
  putchar(' ');
}
inline void add(int x, int y) {
  a[++tot].t = y;
  a[tot].next = head[x];
  head[x] = tot;
}
inline void pre_dfs(int x) {
  dfn[x] = ++cl;
  for (int i = 1; i < 20; i++) fa[x][i] = fa[fa[x][i - 1]][i - 1];
  for (int i = head[x]; i; i = a[i].next) {
    int t = a[i].t;
    if (t == fa[x][0]) continue;
    dep[t] = dep[x] + 1;
    fa[t][0] = x;
    pre_dfs(t);
  }
}
inline int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  int t = dep[x] - dep[y];
  for (int i = 19; ~i; i--)
    if ((t >> i) & 1) x = fa[x][i];
  if (x == y) return x;
  for (int i = 19; ~i; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
inline int getdis(int x, int y) { return dep[x] + dep[y] - dep[lca(x, y)] * 2; }
int st[400010], p[400010], pt, top;
inline bool cmp(const int &x, const int &y) { return dfn[x] < dfn[y]; }
inline void link(int x, int y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
inline void build() {
  sort(c + 1, c + cnt + 1, cmp);
  cnt = unique(c + 1, c + cnt + 1) - c - 1;
  st[top = 1] = 1;
  p[pt = 1] = 1;
  for (int i = 2; i <= cnt; i++) {
    int x = c[i], y = lca(st[top], x);
    p[++pt] = x;
    p[++pt] = y;
    while (top > 1 && dep[st[top - 1]] > dep[y])
      link(st[top - 1], st[top]), top--;
    if (top && dep[y] < dep[st[top]]) link(y, st[top]), top--;
    if (y != st[top]) st[++top] = y;
    st[++top] = x;
  }
  while (top > 1) link(st[top - 1], st[top]), top--;
}
pair<int, int> dis[400010];
int from[400010];
bool inq[400010];
struct hhhh {
  pair<int, int> y;
  int x;
  hhhh(pair<int, int> yy, int xx) : y(yy), x(xx) {}
};
inline bool operator<(const hhhh &x, const hhhh &y) { return y.y < x.y; }
priority_queue<hhhh> q;
inline void work() {
  int m1 = rd(), m2 = rd();
  c[cnt = 1] = 1;
  for (int i = 1; i <= m1; i++)
    h[i].first = rd(), h[i].second = rd(), c[++cnt] = h[i].first;
  for (int i = 1; i <= m2; i++) num[i] = rd(), c[++cnt] = num[i];
  build();
  while (!q.empty()) q.pop();
  for (int i = 1; i <= pt; i++)
    from[p[i]] = 0, dis[p[i]] = pair<int, int>(n + 1, 0), inq[p[i]] = false;
  for (int i = 1; i <= m1; i++)
    from[h[i].first] = i, dis[h[i].first] = pair<int, int>(0, i),
    q.push(hhhh(pair<int, int>(0, i), h[i].first));
  while (!q.empty()) {
    hhhh hh = q.top();
    q.pop();
    int x = hh.x, y = from[x];
    if (inq[x]) continue;
    inq[x] = true;
    for (int t : v[x]) {
      if (inq[t]) continue;
      pair<int, int> now(
          (getdis(t, h[y].first) + h[y].second - 1) / h[y].second, y);
      if (now < dis[t])
        dis[t] = now, from[t] = now.second, q.push(hhhh(now, t));
    }
  }
  for (int i = 1; i <= m2; i++) print(from[num[i]]);
  putchar('\n');
  for (int i = 1; i <= pt; i++) v[p[i]].clear();
}
int main() {
  n = rd();
  tot = 0;
  for (int i = 1; i < n; i++) {
    int x = rd(), y = rd();
    add(x, y);
    add(y, x);
  }
  cl = 0;
  pre_dfs(1);
  m = rd();
  while (m--) work();
  return 0;
}