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;
  const int maxN = 2e5 + 10;
  vector<int> g[maxN];
  const int BUBEN = 400;
  int q;
  long long ans[maxN];
  vector<int> st[BUBEN + 2];
  int k[maxN], vv[maxN];
  long long dp[maxN];
  vector<int> ord;
  vector<int> edge[maxN];
  int par[maxN];
  int tin[maxN];
  int timer = 0;
  int tout[maxN];
  int h[maxN];
  void dfs(int v, int p) {
    tin[v] = timer++;
    for (int to : g[v]) {
      if (to == p) continue;
      par[to] = v;
      edge[v].emplace_back(to);
      h[to] = h[v] + 1;
      dfs(to, v);
    }
    ord.emplace_back(v);
    tout[v] = timer++;
  }
  int bigPar[maxN];
  vector<pair<int, int>> big_edge[maxN];
  bool byTin(int a, int b) { return tin[a] < tin[b]; }
  int lca(int x, int y) {
    while (h[x] > h[y]) {
      x = par[x];
    }
    while (h[x] < h[y]) {
      y = par[y];
    }
    while (x != y) {
      x = par[x];
      y = par[y];
    }
    return x;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      g[x].emplace_back(y);
      g[y].emplace_back(x);
    }
    dfs(1, -1);
    cin >> q;
    for (int i = 1; i <= q; i++) {
      cin >> vv[i] >> k[i];
      if (k[i] <= BUBEN) {
        st[k[i]].emplace_back(i);
      }
    }
    for (int i = 0; i <= BUBEN; i++) {
      if (!st[i].empty()) {
        for (int vert : ord) {
          dp[vert] = 0;
          for (int x : edge[vert]) {
            dp[vert] += max(1LL, dp[x] - i);
          }
        }
        for (int id : st[i]) {
          ans[id] = dp[vv[id]];
        }
      }
    }
    vector<int> inter;
    for (int z : ord) {
      if ((int)edge[z].size() > BUBEN / 2 || z == 1) {
        inter.emplace_back(z);
      }
    }
    sort(inter.begin(), inter.end(), byTin);
    int m = inter.size();
    for (int i = 0; i + 1 < m; i++) {
      inter.emplace_back(lca(inter[i], inter[i + 1]));
    }
    sort(inter.begin(), inter.end(), byTin);
    inter.erase(unique(inter.begin(), inter.end()), inter.end());
    vector<int> st;
    st.push_back(inter[0]);
    for (int i = 1; i < inter.size(); i++) {
      int x = inter[i];
      while (tout[st.back()] < tin[x]) st.pop_back();
      int y = st.back();
      big_edge[y].emplace_back(make_pair(x, -h[y] + h[x]));
      bigPar[x] = y;
      st.push_back(x);
    }
    reverse(inter.begin(), inter.end());
    for (int i = 1; i <= q; i++) {
      if (k[i] > BUBEN) {
        int who = vv[i];
        vector<int> without_par;
        bool hs = false;
        for (int u : inter) {
          if (u == who) hs = true;
          if (tin[who] <= tin[u] && tout[u] <= tout[who]) {
            if (u == 1 ||
                !(tin[who] <= tin[bigPar[u]] && tout[bigPar[u]] <= tout[who])) {
              without_par.emplace_back(u);
            }
            dp[u] = 0;
            for (auto& it : big_edge[u]) {
              dp[u] += max(1LL, dp[it.first] - 1LL * it.second * k[i]);
            }
            dp[u] += edge[u].size() - big_edge[u].size();
          }
        }
        if (hs) {
          cout << dp[who] << '\n';
        } else {
          dp[who] = 0;
          for (int z : without_par) {
            dp[who] += max(1LL, dp[z] - 1LL * k[i] * (h[z] - h[who]));
          }
          dp[who] += edge[who].size() - without_par.size();
          cout << dp[who] << '\n';
        }
      } else {
        cout << ans[i] << '\n';
      }
    }
    return 0;
  }
  