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>
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimize(0)
  #pragma GCC optimization("unroll-loops")
  using namespace std;
  const int MAXN = 1e3 + 1, inf = 1e9 + 7;
  int M = 1e9 + 7, P = 1e6 + 2;
  int f[MAXN + 1], inv[MAXN + 1], inc[MAXN + 1];
  int bin_pow(int a, int b, int M) {
    if (b <= 0) return 1;
    if (b % 2 == 0) {
      int x = bin_pow(a, b / 2, M);
      return x * x % M;
    } else {
      return bin_pow(a, b - 1, M) * a % M;
    }
  }
  int C(int n, int k) {
    if (n < 0 || k > n) return 0;
    return (f[n] * inc[k] % M * inc[n - k] % M + M) % M;
  }
  int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
  int lcm(int a, int b) {
    if (a < 0) return lcm(abs(a), b);
    if (b < 0) return lcm(a, abs(b));
    if (a == 0 && b == 0) return 0;
    return a / gcd(a, b) * b;
  }
  vector<vector<int>> g, dp;
  int get(int u, int k) {
    k = min(k, (int)dp[u].size() - 1);
    return dp[u][k];
  }
  void dfs(int u, int p) {
    int cnt = 0;
    vector<int> to;
    for (int v : g[u]) {
      if (v != p) {
        dfs(v, u);
        cnt++;
        to.push_back(v);
      }
    }
    int s = 0;
    for (int i = 0; i < (int)g.size(); ++i) {
      int tmp = 0;
      for (int v : to) {
        tmp += max((int)1, get(v, i) - i);
      }
      dp[u].push_back(tmp + s);
      if (i == 250) {
        vector<int> to_2;
        for (int v : to) {
          if (get(v, i) - i < 2)
            ++s;
          else
            to_2.push_back(v);
        }
        swap(to, to_2);
      }
      if (tmp + s == cnt) break;
    }
  }
  void solve() {
    int n, m;
    cin >> n;
    g.resize(n);
    dp.resize(n);
    for (int i = 0; i < n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      --u;
      --v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    dfs(0, -1);
    int q;
    cin >> q;
    while (q--) {
      int u, k;
      cin >> u >> k;
      --u;
      cout << get(u, k) << '\n';
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    f[0] = inv[0] = f[1] = inv[1] = 1;
    for (int i = 2; i <= MAXN; ++i) {
      f[i] = f[i - 1] * i % M;
      inv[i] = ((-M / i) * inv[M % i] + M) % M;
    }
    inc[0] = 1;
    for (int i = 1; i <= MAXN; ++i) {
      inc[i] = inc[i - 1] * inv[i] % M;
    }
    int _ = 1;
    while (_--) solve();
  }
  