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 + 3;
  basic_string<int> g[N], qr[N], f[N];
  int a[N], d[N], ans[N];
  struct Q {
    int v, k;
  } o[N];
  void dfs1(int x, int y) {
    a[x] = -1, d[x] = d[y] + 1;
    for (int i : g[x])
      if (i ^ y) dfs1(i, x), ++a[x];
  }
  void dp(int x, int y) {
    if (a[x] < 0) {
      return;
    }
    int l, j;
    for (int i : g[x])
      if (i ^ y) {
        dp(i, x);
        if (f[x].size() < f[i].size()) f[x].resize(f[i].size());
        l = f[i].size();
        for (j = 0; j < l; ++j) f[x][j] += f[i][j];
      }
    for (int i : qr[x])
      if (o[i].k < f[x].size()) ans[i] = f[x][o[i].k];
    if (f[x].size() < a[x]) f[x].resize(a[x]);
    l = f[x].size();
    for (j = 0; j < l; ++j) f[x][j] += a[x] - j;
    while (f[x].back() < 0) f[x].pop_back();
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n, q, i, j, k;
    cin >> n;
    for (i = 1; i < n; ++i) {
      cin >> j >> k;
      g[j] += k, g[k] += j;
    }
    dfs1(1, 0);
    cin >> q;
    for (i = 1; i <= q; ++i) {
      cin >> o[i].v >> o[i].k;
      qr[o[i].v] += i;
    }
    dp(1, 0);
    for (i = 1; i <= q; ++i) cout << ans[i] + a[o[i].v] + 1 << '\n';
    return 0;
  }
  