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;
  vector<int> L[200200];
  int ord[2][200200];
  int cnt = 0;
  int par[200200];
  int sum_kids[200200];
  int dp[200200];
  vector<pair<int, int> > queries[200200];
  void dfs(int vx, int p = 0) {
    par[vx] = p;
    for (int i : L[vx])
      if (i != p) {
        dfs(i, vx);
        dp[vx] += dp[i];
      }
    ord[0][cnt++] = vx;
    dp[vx] = max(dp[vx], 1);
    sum_kids[p] += dp[vx];
  }
  int ans[200200];
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      L[a].push_back(b);
      L[b].push_back(a);
    }
    int q;
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
      int v, k;
      scanf("%d%d", &v, &k);
      queries[k].push_back({v, i});
    }
    dfs(1);
    int sz = n;
    for (int k = 1; k < 200200; k++) {
      for (pair<int, int> i : queries[k - 1]) ans[i.second] = sum_kids[i.first];
      int nu = 0;
      for (int i = 0; i < sz; i++) {
        int vx = ord[(k + 1) % 2][i];
        sum_kids[par[vx]] -= dp[vx];
        dp[vx] = max(1, sum_kids[vx] - k);
        sum_kids[par[vx]] += dp[vx];
        if (dp[vx] > 1) ord[k % 2][nu++] = vx;
      }
      sz = nu;
    }
    for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
  }
  