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;
  struct frac {
    long long b, a;
    frac(long long b, long long a) : b(b), a(a) {}
    bool operator<(const frac &p) const { return b * p.a < p.b * a; }
  };
  struct fenwick {
    long long fen[200005];
    void update(int i, long long x) {
      while (i <= 200000) {
        fen[i] += x;
        i += i & -i;
      }
    }
    long long query(int i) const {
      long long r = 0;
      while (i) {
        r += fen[i];
        i -= i & -i;
      }
      return r;
    }
    long long query(int x, int y) const { return query(y) - query(x - 1); }
  } fen0, fen1;
  int n, q;
  vector<int> edge[200005];
  int st[200005], ed[200005];
  int child[200005];
  vector<pair<frac, int>> L;
  void dfs(int x, int p, priority_queue<frac> &pq) {
    static int ord = 0;
    st[x] = ++ord;
    for (int i : edge[x]) {
      if (i == p) continue;
      ++child[x];
      priority_queue<frac> c;
      dfs(i, x, c);
      if (pq.size() < c.size()) swap(pq, c);
      while (!c.empty()) {
        pq.push(c.top());
        c.pop();
      }
    }
    frac f = frac(child[x] - 1, 1);
    while (!pq.empty() && f < pq.top()) {
      f.b += pq.top().b;
      f.a += pq.top().a;
      L.emplace_back(pq.top(), -p);
      pq.pop();
    }
    pq.push(f);
    L.emplace_back(f, p);
    ed[x] = ord;
  }
  long long ans[200005];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i < n; ++i) {
      int x, y;
      cin >> x >> y;
      edge[x].push_back(y);
      edge[y].push_back(x);
    }
    priority_queue<frac> c;
    dfs(1, 0, c);
    sort(L.begin(), L.end());
    cin >> q;
    vector<tuple<int, int, int>> Q;
    for (int i = 1; i <= q; ++i) {
      int v, k;
      cin >> v >> k;
      Q.emplace_back(k, v, i);
    }
    sort(Q.rbegin(), Q.rend());
    for (auto [k, v, i] : Q) {
      while (!L.empty() && frac(k, 1) < L.back().first) {
        auto [f, x] = L.back();
        L.pop_back();
        if (x < 0) {
          fen0.update(st[-x], -f.b);
          fen1.update(st[-x], -f.a);
        } else if (x > 0) {
          fen0.update(st[x], f.b);
          fen1.update(st[x], f.a);
        }
      }
      ans[i] = child[v] + fen0.query(st[v], ed[v]) - fen1.query(st[v], ed[v]) * k;
    }
    for (int i = 1; i <= q; ++i) {
      printf("%lld\n", ans[i]);
    }
    return 0;
  }
  