You are given a tree consisting of n vertices. Recall that a tree is an undirected connected acyclic graph. The given tree is rooted at the vertex 1.

You have to process q queries. In each query, you are given a vertex of the tree v and an integer k.

To process a query, you may delete any vertices from the tree in any order, except for the root and the vertex v. When a vertex is deleted, its children become the children of its parent. You have to process a query in such a way that maximizes the value of c(v) - m ⋅ k (where c(v) is the resulting number of children of the vertex v, and m is the number of vertices you have deleted). Print the maximum possible value you can obtain.

The queries are independent: the changes you make to the tree while processing a query don't affect the tree in other queries.

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

Then n-1 lines follow, the i-th of them contains two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the endpoints of the i-th edge. These edges form a tree.

The next line contains one integer q (1 ≤ q ≤ 2 ⋅ 10^5) — the number of queries.

Then q lines follow, the j-th of them contains two integers v_j and k_j (1 ≤ v_j ≤ n; 0 ≤ k_j ≤ 2 ⋅ 10^5) — the parameters of the j-th query.

Output

For each query, print one integer — the maximum value of c(v) - m ⋅ k you can achieve.

Example

Input


8
6 7
3 2
8 3
5 7
7 4
7 1
7 3
6
1 0
1 2
1 3
7 1
5 0
7 200000


Output


5
2
1
4
0
4

Note

The tree in the first example is shown in the following picture:

<image>

Answers to the queries are obtained as follows:

  1. v=1,k=0: you can delete vertices 7 and 3, so the vertex 1 has 5 children (vertices 2, 4, 5, 6, and 8), and the score is 5 - 2 ⋅ 0 = 5; 
  2. v=1,k=2: you can delete the vertex 7, so the vertex 1 has 4 children (vertices 3, 4, 5, and 6), and the score is 4 - 1 ⋅ 2 = 2. 
  3. v=1,k=3: you shouldn't delete any vertices, so the vertex 1 has only one child (vertex 7), and the score is 1 - 0 ⋅ 3 = 1; 
  4. v=7,k=1: you can delete the vertex 3, so the vertex 7 has 5 children (vertices 2, 4, 5, 6, and 8), and the score is 5 - 1 ⋅ 1 = 4; 
  5. v=5,k=0: no matter what you do, the vertex 5 will have no children, so the score is 0; 
  6. v=7,k=200000: you shouldn't delete any vertices, so the vertex 7 has 4 children (vertices 3, 4, 5, and 6), and the score is 4 - 0 ⋅ 200000 = 4. 
#include <bits/stdc++.h>
long long read() {
  long long a = 0;
  int op = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') op = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    a = (a << 3) + (a << 1) + (48 ^ ch);
    ch = getchar();
  }
  return a * op;
}
const int B1 = 2000;
const int B2 = 100;
const int N = 2e5 + 10;
std::vector<int> tr[N];
using dp_t = int*;
int ans[N];
std::vector<std::pair<int, int>> mount[N];
int size[N];
std::pair<dp_t, dp_t> dfs(int now, int from) {
  size[now] = 1;
  dp_t dp1 = nullptr;
  dp_t dp2 = nullptr;
  int d = tr[now].size() - (now != from);
  for (int to : tr[now])
    if (to != from) {
      auto [to1, to2] = dfs(to, now);
      if (dp1 == nullptr) {
        dp1 = new int[B1];
        dp2 = new int[B2];
        for (int i = 0; i < B1; i++) dp1[i] = d;
        for (int i = 0; i < B2; i++) dp2[i] = 0;
      }
      for (int i = 0; i < B1; i++)
        if (to1[i] > i) dp1[i] += to1[i] - 1 - i;
      for (int i = std::min(B2 - 1, size[now]); ~i; i--) {
        dp2[i]++;
        for (int j = 0; j < i && j <= size[to]; j++) {
          dp2[i] = std::max(dp2[i], dp2[i - j - 1] + to2[j]);
        }
      }
      delete to1;
      delete to2;
      size[now] += size[to];
    }
  if (dp1 == nullptr) {
    dp1 = new int[B1];
    dp2 = new int[B2];
    for (int i = 0; i < B1; i++) dp1[i] = d;
    for (int i = 0; i < B2; i++) dp2[i] = 0;
  }
  for (auto [k, id] : mount[now]) {
    if (k < B1)
      ans[id] = dp1[k];
    else
      for (int i = 0; i < B2; i++) ans[id] = std::max(ans[id], dp2[i] - k * i);
  }
  return {dp1, dp2};
}
int main() {
  int n = read();
  for (int i = 1; i < n; i++) {
    int u = read();
    int v = read();
    tr[u].push_back(v);
    tr[v].push_back(u);
  }
  int m = read();
  for (int i = 1; i <= m; i++) {
    int v = read();
    int k = read();
    mount[v].emplace_back(k, i);
  }
  dfs(1, 1);
  for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
  return 0;
}
