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;
  long long cc = 0;
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n;
    cin >> n;
    vector<vector<int>> g(n);
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      --u, --v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    vector<vector<int>> dp(n);
    vector<int> sz(n);
    function<void(int, int)> dfs = [&](int u, int fa) {
      sz[u] = 1;
      int son = 0;
      vector<int> res;
      for (int v : g[u]) {
        if (v == fa) continue;
        dfs(v, u);
        sz[u] += sz[v];
        son += 1;
        for (int k = 0;; ++k) {
          if (k == dp[u].size()) dp[u].push_back(0);
          int p = k < dp[v].size() ? dp[v][k] - k : dp[v].back() - k;
          if (p <= 0) {
            res.push_back(k);
            break;
          }
          dp[u][k] += p;
        }
      }
      if (dp[u].empty()) dp[u].push_back(0);
      int m = dp[u].size();
      cc += m;
      vector<int> c(m + 1);
      for (int i : res) c[i] += 1;
      for (int i = 1; i < m; ++i) c[i] += c[i - 1];
      for (int i = 0; i < m; ++i) dp[u][i] += c[i];
    };
    assert(cc < n + 5);
    dfs(0, -1);
    int q;
    cin >> q;
    while (q--) {
      int u, k;
      cin >> u >> k;
      --u;
      if (k < dp[u].size())
        cout << dp[u][k] << '\n';
      else
        cout << dp[u].back() << '\n';
    }
  }
  