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;
int n, num, tot, tot2, q, m, k, top, h[400005], fir[400005], fir2[400005],
    son[400005 << 1], son2[400005 << 1], nxt[400005 << 1], nxt2[400005 << 1];
int ans[400005], f[400005][20], dep[400005], s[400005], dfn[400005], st[400005];
bool vis[400005];
struct node {
  int h;
} p[400005];
struct node2 {
  int h, co, t, s, v;
  bool operator<(const node2& a) const {
    if (a.t != t) return t > a.t;
    return co > a.co;
  }
} pp[400005];
priority_queue<node2> heap;
inline void add(int x, int y) {
  son[++tot] = y, nxt[tot] = fir[x], fir[x] = tot;
}
inline void add2(int x, int y) {
  son2[++tot2] = y, nxt2[tot2] = fir2[x], fir2[x] = tot2;
}
inline void dfs1(int x) {
  dep[x] = dep[f[x][0]] + 1;
  s[x] = 1;
  for (register int i = 1; i <= 18; i++) f[x][i] = f[f[x][i - 1]][i - 1];
  for (register int i = fir[x]; i; i = nxt[i])
    if (son[i] != f[x][0]) f[son[i]][0] = x, dfs1(son[i]), s[x] += s[son[i]];
}
inline void dfs2(int x) {
  dfn[x] = ++num;
  for (register int i = fir[x]; i; i = nxt[i])
    if (son[i] != f[x][0]) dfs2(son[i]);
}
inline int LCA(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (register int i = 18; i >= 0; i--)
    if (dep[f[x][i]] >= dep[y]) x = f[x][i];
  if (x == y) return x;
  for (register int i = 18; i >= 0; i--)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
inline int dis(int x, int y) { return dep[x] + dep[y] - 2 * dep[LCA(x, y)]; }
inline void mktree(int x) {
  if (top <= 1) {
    if (x != 1) st[++top] = x;
    return;
  }
  int lca = LCA(x, st[top]);
  if (lca == st[top]) {
    st[++top] = x;
    return;
  }
  while (top > 1 && dep[st[top - 1]] >= dep[lca])
    add2(st[top - 1], st[top]), add2(st[top], st[top - 1]), top--;
  if (st[top] != lca) add2(st[top], lca), add2(lca, st[top]), st[top] = lca;
  st[++top] = x;
}
inline void clear(int x, int fa) {
  for (register int i = fir2[x]; i; i = nxt2[i])
    if (son2[i] != fa) clear(son2[i], x);
  ans[x] = fir2[x] = 0, vis[x] = 0;
}
inline bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
inline int read() {
  int ret = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) ret = (ret << 1) + (ret << 3) + ch - '0', ch = getchar();
  return ret * f;
}
int main() {
  n = read();
  for (register int i = 1, x, y; i < n; i++)
    x = read(), y = read(), add(x, y), add(y, x);
  dfs1(1), dfs2(1);
  q = read();
  for (register int i = 1; i <= q; i++) {
    while (!heap.empty()) heap.pop();
    k = read();
    m = read();
    for (register int j = 1; j <= k; j++)
      pp[j].t = 0, pp[j].co = j, pp[j].h = read(), vis[pp[j].h] = 1,
      pp[j].v = pp[j].s = read(), h[j] = pp[j].h, heap.push(pp[j]);
    for (register int j = 1; j <= m; j++) p[j].h = read(), h[k + j] = p[j].h;
    k += m;
    sort(h + 1, h + k + 1, cmp);
    tot2 = 0;
    st[top = 1] = 1;
    for (register int j = 1; j <= k; j++)
      if (h[j] != h[j - 1]) mktree(h[j]);
    while (top > 1)
      add2(st[top], st[top - 1]), add2(st[top - 1], st[top]), top--;
    while (!heap.empty()) {
      node2 tmp = heap.top();
      vis[tmp.h] = 1;
      heap.pop();
      int x = tmp.h, co = tmp.co, tim = tmp.t, s = tmp.s, v = tmp.v;
      if (ans[x]) continue;
      ans[x] = co;
      for (register int i = fir2[x]; i; i = nxt2[i]) {
        int to = son2[i];
        if (vis[to]) continue;
        if (dis(to, x) <= s) {
          heap.push(node2{to, co, tim, s - dis(to, x), v});
          continue;
        }
        int diss = dis(to, x), tt = ceil((double)(diss - s) / (double)v);
        node2 tmpp;
        tmpp.h = to, tmpp.co = tmp.co, tmpp.t = tim + tt,
        tmpp.s = tt * v - (diss - s), tmpp.v = v;
        heap.push(tmpp);
      }
    }
    for (register int j = 1; j <= m; j++) cout << ans[p[j].h] << ' ';
    clear(1, 0);
    puts("");
  }
  return 0;
}