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 N = 2e5 + 5;
  int read() {
    int s = 0;
    char c = getchar(), lc = '+';
    while (c < '0' || '9' < c) lc = c, c = getchar();
    while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
    return lc == '-' ? -s : s;
  }
  void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10)
      putchar(x + '0');
    else
      write(x / 10), putchar(x % 10 + '0');
  }
  void print(int x, char c = '\n') {
    write(x);
    putchar(c);
  }
  vector<int> e[N];
  vector<int> f[N], g[N];
  int deg[N], sum[N];
  void dfs(int now, int father) {
    int mx = 1;
    for (int to : e[now]) {
      if (to == father) continue;
      dfs(to, now);
      deg[now]++;
      mx = max(mx, (int)g[to].size());
    }
    mx = max(mx, deg[now]);
    g[now].resize(mx);
    for (int i = 0; i < mx; i++) g[now][i] = -i;
    for (int to : e[now]) {
      if (to == father) continue;
      for (int i = 0; i < (int)g[to].size(); i++) g[now][i] += g[to][i];
      sum[g[to].size()]++;
    }
    for (int i = 0; i < mx; i++) g[now][i] += sum[i] += i ? sum[i - 1] : 0;
    for (int i = 0; i < mx; i++) f[now].push_back(g[now][i] + i);
    while (!g[now].empty() && g[now].back() < 1) g[now].pop_back();
    for (int i = 0; i <= mx; i++) sum[i] = 0;
  }
  void Multi_test_no_horses() {
    int n = read();
    for (int i = 1; i < n; i++) {
      int u = read(), v = read();
      e[u].push_back(v);
      e[v].push_back(u);
    }
    dfs(1, 0);
    int Q = read();
    while (Q--) {
      int x = read(), k = read();
      if (k >= (int)f[x].size())
        print(deg[x]);
      else
        print(max(deg[x], f[x][k]));
    }
  }
  signed main(signed Hello, char *CodeForces[]) {
    (void)Hello, (void)CodeForces;
    int T = 1;
    while (T--) Multi_test_no_horses();
    return 0;
  }
  