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;
  vector<int> get_parents(const vector<vector<int>>& adj, int root) {
    int n = adj.size();
    vector<int> parents(n, -1);
    stack<pair<int, int>> st;
    st.emplace(root, -1);
    while (!st.empty()) {
      int node = st.top().first;
      parents[node] = st.top().second;
      st.pop();
      for (int i : adj[node]) {
        if (i != parents[node]) {
          st.emplace(i, node);
        }
      }
    }
    return parents;
  }
  vector<vector<int>> get_children(const vector<int>& parents) {
    int n = parents.size();
    vector<vector<int>> children(n);
    for (int i = 0; i < n; ++i) {
      if (parents[i] >= 0) {
        children[parents[i]].push_back(i);
      }
    }
    return children;
  }
  vector<int> get_depth(const vector<vector<int>>& children, int root) {
    int n = children.size();
    vector<int> depth(n);
    vector<int> q(1, root);
    int d = 0;
    while (!q.empty()) {
      vector<int> q2;
      for (int i : q) {
        depth[i] = d;
        for (int j : children[i]) {
          q2.push_back(j);
        }
      }
      q2.swap(q);
      d++;
    }
    return depth;
  }
  vector<int> get_order(const vector<vector<int>>& children, int root) {
    int N = children.size();
    vector<int> order(N);
    int idx = 0;
    stack<pair<bool, int>> st;
    st.emplace(true, root);
    while (!st.empty()) {
      bool b = st.top().first;
      int node = st.top().second;
      st.pop();
      if (b) {
        st.emplace(false, node);
        for (int ch : children[node]) {
          st.emplace(true, ch);
        }
      } else {
        order[idx++] = node;
      }
    }
    assert(idx == N);
    return order;
  }
  vector<int> calculate_small_dp(const vector<vector<int>>& children,
                                 const vector<int>& order, int k) {
    int n = children.size();
    vector<int> dp(n, 0);
    for (int i : order) {
      for (int ch : children[i]) {
        dp[i] += max(dp[ch] - k, 1);
      }
    }
    return dp;
  }
  int& chmax(int& x, int y) {
    if (y > x) x = y;
    return x;
  }
  vector<int> solve(const vector<vector<int>>& adj, const vector<int>& v,
                    const vector<int>& k) {
    const int THRESH1 = 400;
    const int THRESH2 = 500;
    int n = adj.size();
    vector<int> parents = get_parents(adj, 0);
    vector<vector<int>> children = get_children(parents);
    vector<int> order = get_order(children, 0);
    int q = v.size();
    vector<int> ans(q);
    vector<vector<int>> by_k(THRESH1);
    for (int qi = 0; qi < q; ++qi) {
      if (k[qi] < THRESH1) {
        by_k[k[qi]].push_back(qi);
      }
    }
    for (int ki = 0; ki < THRESH1; ++ki) {
      if (by_k[ki].empty()) continue;
      vector<int> dp = calculate_small_dp(children, order, ki);
      for (int qi : by_k[ki]) {
        ans[qi] = dp[v[qi]];
      }
    }
    vector<vector<int>> dp(n);
    for (int i : order) {
      dp[i].resize(1, 0);
      for (int ch : children[i]) {
        int m = min(int(dp[i].size()) + int(dp[ch].size()), THRESH2 + 1);
        vector<int> u(m, 0);
        for (int j = 0; j < int(dp[i].size()); ++j) {
          u[j] = dp[i][j] + 1;
        }
        for (int j1 = 0; j1 < int(dp[i].size()); ++j1) {
          for (int j2 = 0; (j2 < int(dp[ch].size())) && ((j1 + j2 + 1) < m);
               ++j2) {
            chmax(u[j1 + 1 + j2], dp[i][j1] + dp[ch][j2]);
          }
        }
        u.swap(dp[i]);
      }
    }
    for (int qi = 0; qi < q; ++qi) {
      if (k[qi] >= THRESH1) {
        for (int i = 0; i < int(dp[v[qi]].size()); ++i) {
          chmax(ans[qi], dp[v[qi]][i] - i * k[qi]);
        }
      }
    }
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    vector<vector<int>> adj(n);
    for (int i = 0; i < (n - 1); ++i) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    int q;
    cin >> q;
    vector<int> v(q), k(q);
    for (int i = 0; i < q; ++i) {
      cin >> v[i] >> k[i];
      --v[i];
    }
    vector<int> ans = solve(adj, v, k);
    for (int i = 0; i < q; ++i) {
      cout << ans[i] << "\n";
    }
  }
  