description: 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. 
solution: #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 = 110;
  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 = B2 - 1; ~i; i--) {
          dp2[i]++;
          for (int j = std::max(0, i - size[now] - 1); j < i && j <= size[to];
               j++) {
            dp2[i] = std::max(dp2[i], dp2[i - j - 1] + to2[j]);
          }
        }
        size[now] += size[to];
        delete to1;
        delete to2;
      }
    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;
  }
  