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>
  using namespace std;
  using ll = long long;
  using cmp = complex<int>;
  const int N = 2e5 + 5;
  int n, q, ans[N];
  vector<int> G[N];
  vector<pair<int, int>> Q[N];
  vector<cmp> f[N];
  int slp(const cmp& a) { return imag(a) / real(a); }
  void dfs(int u, int fa) {
    for (int v : G[u])
      if (v != fa) {
        dfs(v, u);
        vector<cmp> t;
        swap(t, f[u]);
        for (int i = 0, j = 0; i < int((t).size()) || j < int((f[v]).size());)
          if (j == int((f[v]).size()) ||
              (i < int((t).size()) && slp(t[i]) <= slp(f[v][j])))
            f[u].push_back(t[i++]);
          else if (i && slp(f[u].back()) == slp(f[v][j]))
            f[u].back() += f[v][j++];
          else
            f[u].push_back(f[v][j++]);
      }
    for (auto [k, i] : Q[u]) {
      cmp sum;
      for (const cmp& x : f[u])
        if (slp(x) >= k) sum += x;
      ans[i] = imag(sum) + int((G[u]).size()) - (u > 1) - k * real(sum);
    }
    cmp t(1, int((G[u]).size()) - (u > 1) - 1);
    while (int((f[u]).size()) && slp(t) <= slp(f[u].back()))
      t += f[u].back(), f[u].pop_back();
    if (imag(t) > 0) f[u].push_back(t);
  }
  int main() {
    cin >> n;
    int u, v;
    for (int i = (2); i <= (n); i++)
      scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
    cin >> q;
    for (int i = (1); i <= (q); i++)
      scanf("%d%d", &u, &v), Q[u].emplace_back(v, i);
    dfs(1, 0);
    for (int i = (1); i <= (q); i++) printf("%d\n", ans[i]);
  }
  