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, q, h[200005], cnt, a[200005], v[200005], b[200005], deep[200005],
    sta[200005], tail = 1, p[200005][20], dps[400005], vh[200005], c[400005],
                 dfn[400005], sign, tot, vh2[200005], ans[200005], anss[200005],
                 vfa[200005];
struct Edge {
  int to, next;
} w[400005];
bool cmp(int a, int b) { return dfn[a] < dfn[b]; }
void add(int x, int y) {
  w[++cnt] = (Edge){y, h[x]};
  h[x] = cnt;
}
void dfs(int x, int fa) {
  dfn[x] = ++sign;
  p[x][0] = fa;
  for (int i = 1; i <= 19; i++) p[x][i] = p[p[x][i - 1]][i - 1];
  deep[x] = deep[fa] + 1;
  for (int i = h[x]; i; i = w[i].next) {
    int to = w[i].to;
    if (to == fa) continue;
    dfs(to, x);
  }
}
int lca(int x, int y) {
  if (deep[x] < deep[y]) swap(x, y);
  for (int i = 19; i >= 0; i--)
    if (deep[x] - (1 << i) >= deep[y]) x = p[x][i];
  if (x == y) return x;
  for (int i = 19; i >= 0; i--)
    if (p[x][i] != p[y][i]) x = p[x][i], y = p[y][i];
  return p[x][0];
}
void dp(int x) {
  ans[x] = 0;
  if (vh[x]) ans[x] = vh[x];
  vfa[x] = x;
  for (int i = h[x]; i; i = w[i].next) {
    int to = w[i].to;
    dp(to);
    if (vh[x]) continue;
    if (!ans[x])
      ans[x] = ans[to];
    else {
      if (!ans[to]) continue;
      int t = (deep[a[ans[to]]] - deep[x] - 1) / v[ans[to]] + 1;
      int t2 = (deep[a[ans[x]]] - deep[x] - 1) / v[ans[x]] + 1;
      if (t < t2)
        ans[x] = ans[to];
      else if (t == t2 && ans[to] < ans[x])
        ans[x] = ans[to];
    }
  }
}
void dp2(int x) {
  for (int i = h[x]; i; i = w[i].next) {
    int to = w[i].to;
    if (!vh[to] && ans[x] != ans[to]) {
      if (!ans[to])
        ans[to] = ans[x], vfa[to] = vfa[x];
      else {
        int len = deep[to] + deep[a[ans[x]]] - 2 * deep[vfa[x]];
        int t = (len - 1) / v[ans[x]] + 1;
        int len2 = deep[a[ans[to]]] - deep[to];
        int t2 = (len2 - 1) / v[ans[to]] + 1;
        if (t < t2)
          ans[to] = ans[x], vfa[to] = vfa[x];
        else if (t == t2 && ans[x] < ans[to])
          ans[to] = ans[x], vfa[to] = vfa[x];
      }
    }
    dp2(to);
  }
  if (vh2[x]) anss[vh2[x]] = ans[x];
  h[x] = vh[x] = vh2[x] = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
  }
  dfs(1, 0);
  memset(h, 0, sizeof(h));
  cnt = 0;
  scanf("%d", &q);
  while (q--) {
    int k, m;
    scanf("%d%d", &k, &m);
    cnt = 0;
    tot = 0;
    for (int i = 1; i <= k; i++)
      scanf("%d%d", &a[i], &v[i]), vh[a[i]] = i, c[++tot] = a[i];
    for (int i = 1; i <= m; i++)
      scanf("%d", &b[i]), c[++tot] = b[i], vh2[b[i]] = i;
    sort(c + 1, c + tot + 1);
    tot = unique(c + 1, c + tot + 1) - c - 1;
    sort(c + 1, c + tot + 1, cmp);
    tail = 1;
    sta[tail] = 1;
    for (int i = 1; i <= tot; i++) {
      int x = lca(c[i], sta[tail]);
      while (deep[sta[tail]] > deep[x]) {
        if (deep[sta[tail - 1]] <= deep[x]) {
          add(x, sta[tail]);
          tail--;
          if (sta[tail] != x) sta[++tail] = x;
          break;
        } else
          add(sta[tail - 1], sta[tail]), tail--;
      }
      if (sta[tail] != c[i]) sta[++tail] = c[i];
    }
    while (tail > 1) add(sta[tail - 1], sta[tail]), tail--;
    dp(1);
    dp2(1);
    for (int i = 1; i <= m; i++) printf("%d ", anss[i]);
    puts("");
  }
  return 0;
}