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>
  std::priority_queue<std::pair<long long, int>,
                      std::vector<std::pair<long long, int>>,
                      std::greater<std::pair<long long, int>>>
      pq, del;
  struct node {
    long long a, b;
    node(long long x = 0, long long y = 0) : a(y), b(x) {}
    node operator+(node const &y) const { return node(b + y.b, a + y.a); }
    node operator-(node const &y) const { return node(b - y.b, a - y.a); }
    node operator-() const { return node(-b, -a); }
    long long operator()(long long x) const { return a * x + b; }
    long long zero() const { return std::ceil((1. - b) / (a + 1.)); }
  };
  std::vector<int> v[200010];
  node a[200010], c[200010];
  int bg[200010], f[200010], ct, cnt, ed[200010], q[200010], n, m, fa[200010];
  long long ans[1000010];
  void upd(int i, node p) {
    if (i)
      for (; i <= n; i += i & -i) c[i] = c[i] + p;
  }
  node que(int i) {
    node ans;
    for (; i; i &= i - 1) ans = ans + c[i];
    return ans;
  }
  node query(int x) { return que(ed[x]) - que(bg[x] - 1); }
  void update(int x, int y, node v) { upd(bg[fa[x]], -v), upd(bg[fa[y]], v); }
  void dfs(int x, int f) {
    fa[x] = f;
    ::f[x] = x;
    bg[x] = ++cnt;
    for (auto u : v[x])
      if (u != f) ++a[x].b, dfs(u, x);
    ed[x] = cnt;
    upd(bg[fa[x]], node(1)), upd(bg[fa[fa[x]]], node(-1));
  }
  auto find = [](int x) {
    while (x != f[x]) x = f[x] = f[f[x]];
    return x;
  };
  auto clr = []() {
    while (del.size() && pq.top() == del.top()) pq.pop(), del.pop();
  };
  auto top = []() {
    clr();
    return pq.top();
  };
  auto pop = []() {
    clr();
    pq.pop();
  };
  int main() {
    scanf("%d", &n);
    for (int i = 2, x, y; i <= n; i++)
      scanf("%d%d", &x, &y), v[x].push_back(y), v[y].push_back(x);
    dfs(1, 0);
    for (int i = 2; i <= n; i++)
      if (a[i].b) pq.emplace(a[i].zero(), -i);
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
      long long b;
      scanf("%d%lld", q + i, &b);
      pq.emplace(-b, i);
    }
    while (ct < m) {
      auto [x, id] = top();
      pop();
      if (id < 0) {
        id = -id;
        if (id == 1) continue;
        int u = find(fa[id]);
        f[id] = u;
        if (u != 1) del.emplace(query(u).zero(), -u);
        update(u, id, query(id) + node(-1, 1));
        if (u != 1) pq.emplace(query(u).zero(), -u);
      } else {
        ans[id] = query(q[id])(x);
        ++ct;
      }
    }
    for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
    return 0;
  }
  