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;
  const int N = 200010;
  int n, q, Head[N], Next[N << 1], V[N << 1], cnt;
  void add(int u, int v) {
    V[cnt] = v;
    Next[cnt] = Head[u];
    Head[u] = cnt++;
  }
  vector<int> f[N];
  void dfs(int u, int fa) {
    vector<int> res;
    res.clear();
    for (int i = Head[u]; i != -1; i = Next[i]) {
      if (V[i] != fa) {
        dfs(V[i], u);
        for (int k = 0;; k++) {
          if (k == f[u].size()) f[u].push_back(0);
          int p = k < f[V[i]].size() ? f[V[i]][k] - k : f[V[i]].back() - k;
          if (p <= 0) {
            res.push_back(k);
            break;
          }
          f[u][k] += p;
        }
      }
    }
    if (f[u].empty()) f[u].push_back(0);
    vector<int> c(f[u].size() + 1);
    for (int i = 0; i < res.size(); i++) {
      c[res[i]]++;
    }
    for (int i = 0; i < f[u].size(); i++) {
      if (i != 0) c[i] += c[i - 1];
      f[u][i] += c[i];
    }
  }
  int main() {
    memset(Head, -1, sizeof(Head));
    scanf("%d", &n);
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      add(u, v);
      add(v, u);
    }
    dfs(1, 0);
    scanf("%d", &q);
    while (q--) {
      int u, k;
      scanf("%d%d", &u, &k);
      if (k < f[u].size())
        printf("%d\n", f[u][k]);
      else
        printf("%d\n", f[u].back());
    }
  }
  