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;
  struct Vertex {
    int cost;
    int depth;
    int idx;
    Vertex(){};
    Vertex(int cost, int depth, int idx) : cost(cost), depth(depth), idx(idx){};
  };
  bool operator<(const Vertex& a, const Vertex& b) {
    if (a.cost != b.cost) return a.cost > b.cost;
    if (a.depth != b.depth) return a.depth > b.depth;
    return a.idx < b.idx;
  }
  struct DSU {
    int n;
    vector<int> p;
    vector<int> top;
    int get(int x) {
      if (p[x] == x) return x;
      return p[x] = get(p[x]);
    }
    int get_top(int x) { return top[get(x)]; }
    void merge(int x, int par) { p[x] = par; }
    DSU(int k = 0) {
      n = k;
      p.resize(n);
      iota(p.begin(), p.end(), 0);
      top = p;
    };
  };
  const int N = 200043;
  int n;
  vector<int> g[N];
  int p[N], d[N], tin[N], tout[N];
  int qv[N];
  int qk[N];
  int T = 0;
  void dfs(int v = 0, int par = -1) {
    tin[v] = T++;
    p[v] = par;
    if (par == -1)
      d[v] = 0;
    else
      d[v] = d[par] + 1;
    for (auto x : g[v])
      if (x != par) dfs(x, v);
    tout[v] = T;
  }
  pair<long long, long long> tree[4 * N];
  pair<long long, long long> operator+(const pair<long long, long long>& a,
                                       const pair<long long, long long>& b) {
    return make_pair(a.first + b.first, a.second + b.second);
  }
  pair<long long, long long> get(int v, int l, int r, int L, int R) {
    if (L >= R) return {0, 0};
    if (l == L && r == R) return tree[v];
    int m = (l + r) / 2;
    return get(v * 2 + 1, l, m, L, min(R, m)) +
           get(v * 2 + 2, m, r, max(m, L), R);
  }
  void add(int v, int l, int r, int pos, pair<long long, long long> val) {
    if (l == r - 1)
      tree[v] = tree[v] + val;
    else {
      int m = (l + r) / 2;
      if (pos < m)
        add(v * 2 + 1, l, m, pos, val);
      else
        add(v * 2 + 2, m, r, pos, val);
      tree[v] = tree[v] + val;
    }
  }
  pair<long long, long long> get_vertex_value(int v) {
    return get(0, 0, n, tin[v], tout[v]);
  }
  void add_on_path(int x, int y, pair<long long, long long> val) {
    add(0, 0, n, tin[x], val);
    if (p[y] != -1) add(0, 0, n, tin[p[y]], make_pair(-val.first, -val.second));
  }
  int calculate_cost(int v) {
    pair<long long, long long> res = get_vertex_value(v);
    long long k = (res.first + res.second) / (res.second + 1) - 1;
    return k;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      --x;
      --y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    int q;
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
      scanf("%d %d", &qv[i], &qk[i]);
      --qv[i];
    }
    dfs();
    DSU dsu(n);
    vector<long long> ans(q);
    set<Vertex> pq;
    vector<Vertex> curv(n);
    for (int i = 0; i < n; i++) {
      int count_children = g[i].size();
      if (i != 0) count_children--;
      add_on_path(i, i, make_pair((long long)(count_children), 0ll));
      if (i != 0) {
        curv[i] = Vertex(calculate_cost(i), d[i], i);
        pq.insert(curv[i]);
      }
    }
    for (int i = 0; i < q; i++) pq.insert(Vertex(qk[i], -1, i));
    while (!pq.empty()) {
      Vertex z = *pq.begin();
      pq.erase(pq.begin());
      if (z.depth == -1) {
        auto res = get_vertex_value(qv[z.idx]);
        ans[z.idx] = res.first - res.second * z.cost;
      } else {
        int v = z.idx;
        int pv = p[v];
        int newtop = dsu.get_top(pv);
        pair<long long, long long> val = get_vertex_value(v);
        val.second++;
        val.first--;
        if (newtop != 0) pq.erase(curv[newtop]);
        add_on_path(pv, newtop, val);
        if (newtop != 0) curv[newtop].cost = calculate_cost(newtop);
        if (newtop != 0) pq.insert(curv[newtop]);
        dsu.merge(v, pv);
      }
    }
    for (int i = 0; i < q; i++) printf("%lld\n", ans[i]);
  }
  