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;
  namespace Ruri {
  const int infi = 1e9;
  const long long infl = 4e18;
  }  // namespace Ruri
  using namespace Ruri;
  namespace Gadgets {
  inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
  inline long long pos_mod(long long a, long long b) { return (a % b + b) % b; }
  }  // namespace Gadgets
  namespace Read {
  inline int ri() {
    int x;
    scanf("%d", &x);
    return x;
  }
  inline long long rl() {
    long long x;
    scanf("%lld", &x);
    return x;
  }
  inline double rd() {
    double x;
    scanf("%lf", &x);
    return x;
  }
  }  // namespace Read
  namespace DeBug {}
  using namespace Gadgets;
  using namespace Read;
  using namespace DeBug;
  const int MAX_N = 2e5 + 5;
  int n;
  vector<int> g[MAX_N];
  vector<pair<int, int> > query[MAX_N];
  int son[MAX_N];
  int ans[MAX_N];
  vector<int> dp[MAX_N];
  void Dfs(int u, int fa) {
    dp[u].resize(son[u]);
    for (auto v : g[u])
      if (v != fa) {
        Dfs(v, u);
        if ((int)dp[u].size() < (int)dp[v].size())
          dp[u].resize((int)dp[v].size());
        for (int i = 0, bbb = (int)dp[v].size() - 1; i <= bbb; ++i)
          dp[u][i] += dp[v][i] - 1;
      }
    for (auto x : query[u])
      if (x.second < (int)dp[u].size()) ans[x.first] += dp[u][x.second];
    for (int i = 0, bbb = (int)dp[u].size() - 1; i <= bbb; ++i)
      dp[u][i] += son[u] - i;
    while (!dp[u].empty() && dp[u].back() <= 0) dp[u].pop_back();
  }
  int main() {
    n = ri();
    for (int i = 1, bbb = n - 1; i <= bbb; ++i) {
      int u = ri(), v = ri();
      g[u].push_back(v), g[v].push_back(u);
    }
    for (int i = 1, bbb = n; i <= bbb; ++i) son[i] = (int)g[i].size() - (i != 1);
    int q = ri();
    for (int i = 1, bbb = q; i <= bbb; ++i) {
      int v = ri(), k = ri();
      query[v].push_back(pair<int, int>(i, k));
      ans[i] = son[v];
    }
    Dfs(1, 0);
    for (int i = 1, bbb = q; i <= bbb; ++i) printf("%d\n", ans[i]);
    return 0;
  }
  