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 = 2e5 + 11;
  const long long M = 1e6 + 21;
  const long long big = 1e17;
  const long long hsh2 = 1964325029;
  const long long mod = 1e9 + 7;
  const double EPS = 1e-9;
  const long long block = 100;
  const long long shift = 2e3;
  const double pi = acos(-1.0);
  mt19937_64 bruh(chrono::steady_clock::now().time_since_epoch().count());
  long long n, k;
  long long x[N], y[N];
  long long q, answ[N];
  vector<long long> g[N], ng[N], bg[N];
  vector<pair<long long, long long> > que[N];
  long long dp[N], sum[N];
  long long potential[N], sub[N];
  long long tin[N], arr[N];
  void prec(long long v, long long p = 0) {
    tin[v] = ++tin[0];
    arr[tin[v]] = v;
    for (auto u : g[v])
      if (u != p) prec(u, v), bg[v].push_back(u);
  }
  void calc_potentials(long long v, long long p) {
    potential[v] = 1;
    long long second = 0;
    for (auto u : g[v]) {
      if (u != p) calc_potentials(u, v), second += potential[u];
    }
    potential[v] = max(1ll, second - block);
  }
  void build(long long v, long long p) {
    for (auto u : g[v]) {
      if (u == p) continue;
      if (potential[u] >= block)
        ng[v].push_back(u);
      else
        sub[v]++;
      build(u, v);
    }
  }
  void naive(long long v) {
    long long answ = sub[v];
    dp[v] = 1;
    sum[v] = sub[v];
    for (auto u : ng[v]) {
      naive(u);
      sum[v] += dp[u];
    }
    dp[v] = max(1ll, sum[v] - k);
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i < n; i++) {
      long long a, b;
      cin >> a >> b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    cin >> q;
    for (int i = 1; i <= q; i++) {
      long long v, k;
      cin >> v >> k;
      que[k].push_back(make_pair(v, i));
    }
    prec(1, 0);
    for (int j = 0; j <= block; j++) {
      k = j;
      for (int j = n; j >= 1; j--) {
        long long v = arr[j];
        dp[v] = 0;
        sum[v] = 0;
        for (auto u : bg[v]) sum[v] += dp[u];
        dp[v] = max(1ll, sum[v] - k);
      }
      for (auto u : que[j]) answ[u.second] = sum[u.first];
    }
    calc_potentials(1, 0);
    build(1, 0);
    for (int j = block + 1; j <= 200000; j++) {
      for (auto u : que[j]) {
        k = j;
        naive(u.first);
        answ[u.second] = sum[u.first];
      }
    }
    for (int i = 1; i <= q; i++) cout << answ[i] << '\n';
  }
  