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;
  inline char getc() {
    static char buf[1 << 20], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  template <class T>
  void rd(T& x) {
    int f = 0, c;
    while (!isdigit(c = getc())) f ^= !(c ^ 45);
    x = (c & 15);
    while (isdigit(c = getc())) x = x * 10 + (c & 15);
    if (f) x = -x;
  }
  template <class T>
  void pt(T x, int c = -1) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) pt(x / 10);
    putchar(x % 10 + 48);
    if (c != -1) putchar(c);
  }
  template <class T>
  void umax(T& x, const T& y) {
    if (x < y) x = y;
  }
  const int N = 200005, B = 1500, S = N / B + 3;
  int n, K, Q, f[N], g[N], ans[N], dp[N][S];
  vector<int> e[N];
  vector<pair<int, int> > qs[N], QS[N];
  int ord[N], fa[N], ind;
  void dfs0(int k1, int k2) {
    ord[++ind] = k1, fa[k1] = k2;
    for (auto x : e[k1])
      if (x != k2) dfs0(x, k1);
  }
  void dfs2(int k1, int k2) {
    dp[k1][0] = 0;
    for (auto x : e[k1])
      if (x != k2) {
        dfs2(x, k1);
        for (int a = (S - 1); a >= (0); --a)
          for (int b = (S - 1 - a); b >= (0); --b)
            umax(dp[k1][a + b], dp[k1][a] + dp[x][b]);
      }
    for (auto x : QS[k1]) {
      ans[x.second] = ((int)(e[k1]).size()) - (k1 != 1);
      for (int a = (1); a <= (S - 1); ++a) {
        umax(ans[x.second], dp[k1][a] - a * x.first);
      }
    }
    for (int i = (S - 2); i >= (0); --i) dp[k1][i + 1] = dp[k1][i];
    dp[k1][0] = 1;
  }
  int main() {
    memset(dp, 192, sizeof(dp));
    rd(n);
    for (int i = (2); i <= (n); ++i) {
      int k1, k2;
      rd(k1), rd(k2);
      e[k1].push_back(k2), e[k2].push_back(k1);
    }
    rd(Q);
    for (int i = (1); i <= (Q); ++i) {
      int v, k;
      rd(v), rd(k);
      if (k <= B)
        qs[k].emplace_back(v, i);
      else
        QS[v].emplace_back(k, i);
    }
    dfs0(1, 0);
    for (int k = (0); k <= (B); ++k)
      if (((int)(qs[k]).size())) {
        K = k;
        memset(f, 0, sizeof(f[0]) * (n + 1));
        memset(g, 0, sizeof(g[0]) * (n + 1));
        for (int i = (n); i >= (1); --i) {
          f[ord[i]] = max(g[ord[i]] - K, 1);
          g[fa[ord[i]]] += f[ord[i]];
        }
        for (auto x : qs[k]) {
          ans[x.second] = g[x.first];
        }
      }
    dfs2(1, 0);
    for (int i = (1); i <= (Q); ++i) pt(ans[i], '\n');
    return 0;
  }
  