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 sqn = 58, inf = 1e9;
  int n, eu[200069], pr[200069], dp[200069], dp2[200069][69], tmp[69], tmp2[69],
      sq[200069];
  vector<int> al[200069], ql[200069];
  bitset<200069> vtd;
  pair<int, int> qq[200069];
  void dfs(int x) {
    int i, sz = al[x].size(), l;
    vtd[x] = 1;
    n++;
    eu[n] = x;
    for (i = 0; i < sz; i++) {
      l = al[x][i];
      if (!vtd[l]) {
        pr[l] = x;
        dfs(l);
      }
    }
  }
  int main() {
    int t, rr, i, j, r, k, l, sz, pz;
    scanf("%d", &n);
    for (i = 0; i < n - 1; i++) {
      scanf("%d%d", &k, &l);
      al[k].push_back(l);
      al[l].push_back(k);
    }
    n = 0;
    dfs(1);
    scanf("%d", &t);
    for (rr = 1; rr <= t; rr++) {
      scanf("%d%d", &k, &l);
      qq[rr] = {k, l};
      ql[min(l, n / sqn + 1)].push_back(rr);
    }
    for (i = 0; i <= n / sqn; i++) {
      for (j = 1; j <= n; j++) {
        dp[j] = 0;
      }
      for (j = n; j; j--) {
        k = eu[j];
        dp[pr[k]] += max(dp[k] - i, 1);
      }
      sz = ql[i].size();
      for (j = 0; j < sz; j++) {
        pz = ql[i][j];
        k = qq[pz].first;
        sq[pz] = dp[k];
      }
    }
    for (i = n; i; i--) {
      k = eu[i];
      for (j = 0; j <= sqn; j++) {
        if (!j) {
          tmp[j] = 1;
        } else {
          tmp[j] = dp2[k][j - 1];
        }
        tmp2[j] = -inf;
      }
      for (j = 0; j <= sqn; j++) {
        for (r = 0; r <= sqn - j; r++) {
          tmp2[j + r] = max(tmp2[j + r], dp2[pr[k]][j] + tmp[r]);
        }
      }
      for (j = 0; j <= sqn; j++) {
        dp2[pr[k]][j] = tmp2[j];
      }
    }
    sz = ql[n / sqn + 1].size();
    for (i = 0; i < sz; i++) {
      pz = ql[n / sqn + 1][i];
      k = qq[pz].first;
      l = qq[pz].second;
      for (j = 0; j <= sqn; j++) {
        sq[pz] = max(sq[pz], dp2[k][j] - j * l);
      }
    }
    for (rr = 1; rr <= t; rr++) {
      printf("%d\n", sq[rr]);
    }
  }
  