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;
  int n;
  vector<vector<int>> E;
  vector<int> p;
  vector<int> d;
  vector<int> c;
  void dfs(int cur, int par, int depth) {
    d[cur] = depth;
    p[cur] = par;
    for (int i = 0; i < (E[cur].size()); ++i) {
      int to = E[cur][i];
      if (to == par) continue;
      c[cur]++;
      dfs(to, cur, depth + 1);
    }
  }
  int main() {
    cin >> n;
    E.resize(n);
    p.resize(n, -1);
    for (int i = 0; i < (n - 1); ++i) {
      int x, y;
      scanf("%d %d", &x, &y);
      x--;
      y--;
      E[x].push_back(y);
      E[y].push_back(x);
    }
    d.resize(n, 0);
    c.resize(n, 0);
    dfs(0, -1, 0);
    vector<pair<int, int>> L(n);
    for (int i = 0; i < (n); ++i) {
      L[i] = make_pair(c[i], i);
    }
    sort(L.rbegin(), L.rend());
    int q;
    cin >> q;
    vector<vector<pair<int, int>>> qs(200003);
    for (int i = 0; i < (q); ++i) {
      int v, k;
      scanf("%d %d", &v, &k);
      v--;
      qs[k].emplace_back(v, i);
    }
    vector<int> ans(q);
    vector<int> dp(n, 0);
    vector<bool> f(n, false);
    for (int i = 0; i < (200003); ++i) {
      priority_queue<pair<int, int>> Q;
      for (int j = 0; j < (n); ++j) {
        if (L[j].first <= i + 1) break;
        Q.emplace(d[L[j].second], L[j].second);
        f[L[j].second] = true;
      }
      vector<int> x;
      while (Q.size() > 0) {
        pair<int, int> t = Q.top();
        Q.pop();
        dp[t.second] -= i + 1;
        dp[t.second] += c[t.second];
        x.push_back(t.second);
        if (dp[t.second] <= 0) {
          continue;
        }
        x.push_back(t.second);
        int pp = p[t.second];
        if (pp == -1) continue;
        dp[pp] += dp[t.second];
        if (f[pp]) continue;
        f[pp] = true;
        Q.emplace(d[pp], pp);
      }
      for (int j = 0; j < (qs[i].size()); ++j) {
        int v = qs[i][j].first;
        int ind = qs[i][j].second;
        if (f[v] == false) {
          ans[ind] = c[v] - (i + 1);
        } else {
          ans[ind] = max(c[v] - (i + 1), dp[v]);
        }
        ans[ind] += i + 1;
      }
      for (int j = 0; j < (x.size()); ++j) {
        f[x[j]] = false;
        dp[x[j]] = 0;
      }
    }
    for (int i = 0; i < (q); ++i) {
      printf("%d\n", ans[i]);
    }
    return 0;
  }
  