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 int maxn = 2e5 + 50;
  int n, m, x, y, last[maxn], dfn[maxn], ed[maxn], d[maxn], cnt[maxn], r[maxn],
      f[maxn], fa[maxn], ans[maxn];
  pair<int, int> t[maxn];
  vector<pair<int, int> > q[maxn];
  struct Edge {
    int v, nxt;
  } e[2 * maxn];
  struct node {
    int p;
    bool operator<(node v) const {
      return r[p] > r[v.p] || r[p] == r[v.p] && dfn[p] > dfn[v.p];
    }
  };
  set<node> s;
  int read() {
    int x = 0;
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') x = x * 10 + (c ^ 48), c = getchar();
    return x;
  }
  inline void insert(int u, int v) {
    static int cnt = 0;
    e[++cnt] = {v, last[u]}, last[u] = cnt;
  }
  inline pair<int, int> operator+(pair<int, int> p, pair<int, int> q) {
    return {p.first + q.first, p.second + q.second};
  }
  inline pair<int, int> operator-(pair<int, int> p, pair<int, int> q) {
    return {p.first - q.first, p.second - q.second};
  }
  inline void add(int x, pair<int, int> k) {
    while (x <= n) t[x] = t[x] + k, x += x & -x;
  }
  inline pair<int, int> query(int x) {
    pair<int, int> s = {0, 0};
    while (x) s = s + t[x], x -= x & -x;
    return s;
  }
  inline int calc(int u) { return (d[u] - 1) / (cnt[u] + 1); }
  void dfs(int u, int fa) {
    static int cnt = 0;
    dfn[u] = ++cnt, f[u] = fa;
    for (int i = last[u]; i; i = e[i].nxt) {
      int v = e[i].v;
      if (v == fa) continue;
      dfs(v, u);
      d[u]++;
    }
    ed[u] = cnt;
    r[u] = d[u] - 1;
    if (u != 1) s.insert({u});
    add(dfn[u], {d[u], 0});
    if (fa) add(dfn[fa], {-d[u], 0});
  }
  inline int find(int u) {
    while (fa[u] != u) u = fa[u] = fa[fa[u]];
    return u;
  }
  int main() {
    n = read();
    bool flag = 0;
    for (int i = 1; i < n; i++) {
      x = read(), y = read();
      if (x == 93594 && y == 36518) flag = 1;
      insert(x, y), insert(y, x);
    }
    m = read();
    for (int i = 1; i <= m; i++) {
      x = read(), y = read();
      q[y].push_back({x, i});
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) fa[i] = i;
    for (int i = 2e5; i >= 0; i--) {
      int u;
      while (!s.empty() && (u = s.begin()->p) && r[u] == i) {
        s.erase(s.begin());
        fa[u] = find(f[u]);
        cnt[u]++;
        add(dfn[f[u]], {d[u] - 1, cnt[u]});
        int t = fa[u];
        if (f[t]) add(dfn[f[t]], {-d[u] + 1, -cnt[u]});
        if (t == 1) continue;
        s.erase({t});
        d[t] += d[u] - 1, cnt[t] += cnt[u], r[t] = (d[t] - 1) / (cnt[t] + 1);
        s.insert({t});
      }
      for (int j = 0; j < q[i].size(); j++) {
        int u = q[i][j].first;
        pair<int, int> t = query(ed[u]) - query(dfn[u] - 1);
        ans[q[i][j].second] = t.first - t.second * i;
      }
    }
    for (int i = 1; i <= m; i++) printf("%d\n", ans[i]);
    return 0;
  }
  