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 long long N = 2 * 1e5 + 100;
  long long n, Q, id[N], dfn[N], out[N], cnt, ans[N];
  vector<long long> e[N];
  struct node {
    long long a, b;
  };
  bool operator<(node a, node b) { return a.a * b.b < a.b * b.a; }
  node operator+(node a, node b) {
    a.a += b.a;
    a.b += b.b;
    return a;
  }
  priority_queue<node> q[N];
  vector<pair<node, long long> > F;
  inline long long read() {
    long long f = 1, x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = x * 10 + s - '0';
      s = getchar();
    }
    return x * f;
  }
  struct bit {
    long long sh[N];
    inline long long lowbit(long long x) { return x & (-x); }
    inline void change(long long x, long long v) {
      while (x <= n) sh[x] += v, x += lowbit(x);
    }
    inline long long query(long long x) {
      long long ans = 0;
      while (x) ans += sh[x], x -= lowbit(x);
      return ans;
    }
  } A;
  bit B;
  struct QQ {
    long long v, k, id;
  } sh[N];
  bool operator<(QQ a, QQ b) { return a.k > b.k; }
  void dfs(long long x, long long fa) {
    dfn[x] = ++cnt;
    if (x != 1) e[x].erase(find(e[x].begin(), e[x].end(), fa));
    for (long long u : e[x]) {
      dfs(u, x);
      if ((long long)q[id[x]].size() < (long long)q[id[u]].size())
        swap(id[x], id[u]);
      while (!q[id[u]].empty()) q[id[x]].push(q[id[u]].top()), q[id[u]].pop();
    }
    out[x] = cnt;
    node now = (node){(long long)e[x].size() - 1, 1};
    long long r = id[x];
    while (!q[r].empty() && now < q[r].top()) {
      node tmp = q[r].top();
      q[r].pop();
      now = now + tmp;
      F.emplace_back(tmp, -fa);
    }
    q[r].push(now);
    F.emplace_back(now, fa);
  }
  signed main() {
    n = read();
    for (long long i = 1; i < n; i++) {
      long long u = read(), v = read();
      e[u].push_back(v);
      e[v].push_back(u);
    }
    for (long long i = 1; i <= n; i++) id[i] = i;
    dfs(1, 0);
    Q = read();
    for (long long i = 1; i <= Q; i++) sh[i] = (QQ){read(), read(), i};
    sort(sh + 1, sh + 1 + Q);
    sort(F.begin(), F.end());
    for (long long i = 1; i <= Q; i++) {
      long long k = sh[i].k, v = sh[i].v;
      node now = (node){k, 1};
      while (!F.empty() && now < F.back().first) {
        node tmp = F.back().first;
        long long x = F.back().second;
        F.pop_back();
        if (x == 0) continue;
        long long op = 1;
        if (x < 0) x = -x, op = -1;
        A.change(dfn[x], op * tmp.a);
        B.change(dfn[x], op * tmp.b);
      }
      ans[sh[i].id] = A.query(out[v]) - A.query(dfn[v] - 1) -
                      k * (B.query(out[v]) - B.query(dfn[v] - 1));
      ans[sh[i].id] += (long long)e[v].size();
    }
    for (long long i = 1; i <= Q; i++) printf("%lld\n", ans[i]);
  }
  