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 edge {
  int to, nxxt, len;
} e[400002 << 1];
int n, head[400002], cnt = 1, Q, k, m, fa[400002][20], dep[400002], dfn[400002],
                     rev[400002], cn, st[400002], sp[400002], js;
int b[400002], tot, qs[400002], s[400002], rem[400002], rz, col[400002],
    len[400002], rm[400002];
bool vis[400002];
inline void ins(int u, int v) {
  e[cnt] = (edge){v, head[u], dep[v] - dep[u]};
  head[u] = cnt++;
  e[cnt] = (edge){u, head[v], dep[v] - dep[u]};
  head[v] = cnt++;
}
void df5(int te, int la) {
  dfn[te] = ++cn, rev[cn] = te;
  for (int i = head[te]; i; i = e[i].nxxt) {
    int j = e[i].to;
    if (j == la) continue;
    dep[j] = dep[te] + 1, fa[j][0] = te;
    df5(j, te);
  }
}
inline int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = 19; ~i; i--)
    if (dep[fa[x][i]] >= dep[y]) 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];
}
void clfs(int te, int la) {
  for (int i = head[te]; i; i = e[i].nxxt) {
    int j = e[i].to;
    if (j == la) continue;
    clfs(j, te);
    if (col[j]) {
      int nl = len[j] + e[i].len / sp[col[j]] +
               (rm[j] + e[i].len % sp[col[j]]) / sp[col[j]];
      int nr = (rm[j] + e[i].len % sp[col[j]]) % sp[col[j]];
      int lt = nl + (nr != 0), pt = len[te] + (rm[te] != 0);
      if (lt < pt || (lt == pt && col[j] < col[te]))
        col[te] = col[j], len[te] = nl, rm[te] = nr;
    }
  }
}
void clf5(int te, int la) {
  for (int i = head[te]; i; i = e[i].nxxt) {
    int j = e[i].to;
    if (j == la) continue;
    if (col[te]) {
      int nl = len[te] + e[i].len / sp[col[te]] +
               (rm[te] + e[i].len % sp[col[te]]) / sp[col[te]];
      int nr = (rm[te] + e[i].len % sp[col[te]]) % sp[col[te]];
      int lt = nl + (nr != 0), pt = len[j] + (rm[j] != 0);
      if (lt < pt || (lt == pt && col[te] < col[j]))
        col[j] = col[te], len[j] = nl, rm[j] = nr;
    }
    clf5(j, te);
  }
}
bool cmp(int x, int y) { return dfn[x] < dfn[y]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    ins(x, y);
  }
  fa[1][0] = 1;
  df5(1, 1);
  for (int i = 1; i <= 19; i++)
    for (int j = 1; j <= n; j++) fa[j][i] = fa[fa[j][i - 1]][i - 1];
  memset(head, 0, sizeof(head));
  cnt = 1;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &k, &m);
    tot = js = 0;
    for (int i = 1; i <= k; i++) {
      scanf("%d%d", &st[i], &sp[i]);
      if (!vis[st[i]]) b[++js] = st[i], vis[st[i]] = 1;
    }
    for (int i = 1; i <= m; i++) {
      scanf("%d", &qs[i]);
      if (!vis[qs[i]]) b[++js] = qs[i], vis[qs[i]] = 1;
    }
    if (!vis[1]) b[++js] = 1, vis[1] = 1;
    for (int i = 1; i <= js; i++) rem[i] = b[i];
    rz = js;
    sort(b + 1, b + 1 + js, cmp);
    for (int i = 1; i <= js; i++) {
      if (!tot) {
        s[++tot] = b[i];
        continue;
      }
      int gf = lca(s[tot], b[i]);
      if (gf == s[tot]) {
        s[++tot] = b[i];
        continue;
      }
      while (tot > 1) {
        if (dfn[gf] < dfn[s[tot - 1]]) {
          ins(s[tot - 1], s[tot]);
          tot--;
        } else if (gf == s[tot - 1]) {
          ins(s[tot - 1], s[tot]);
          tot--;
          break;
        } else {
          ins(gf, s[tot]);
          s[tot] = gf;
          rem[++rz] = gf;
          break;
        }
      }
      if (s[tot] ^ b[i]) s[++tot] = b[i];
    }
    while (tot > 1) ins(s[tot - 1], s[tot]), tot--;
    for (int i = 1; i <= rz; i++) col[rem[i]] = 0, len[rem[i]] = 1e9;
    for (int i = 1; i <= k; i++) col[st[i]] = i, len[st[i]] = 0, rm[st[i]] = 0;
    clfs(1, 1);
    clf5(1, 1);
    for (int i = 1; i <= m; i++) printf("%d ", col[qs[i]]);
    puts("");
    for (int i = 1; i <= rz; i++) head[rem[i]] = vis[rem[i]] = 0;
    cnt = 1;
  }
}