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 optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
  using namespace std;
  long long gcd(long long i, long long j) {
    if (j == 0)
      return i;
    else
      return gcd(j, i % j);
  }
  template <typename T>
  inline T getint() {
    T val = 0;
    char c;
    bool neg = false;
    while ((c = getchar()) && !(c >= '0' && c <= '9')) {
      neg |= c == '-';
    }
    do {
      val = (val * 10) + c - '0';
    } while ((c = getchar()) && (c >= '0' && c <= '9'));
    return val * (neg ? -1 : 1);
  }
  const long long INF = 1e9 + 100;
  const int mod = 1000000007;
  const long double eps = 1e-11, pi = acos(-1);
  const long long maxN = 200010, maxT = 1010, A = 179, K = 200;
  mt19937 mt_rand(time(0));
  long long bp(long long et, long long b) {
    b %= mod - 1;
    et %= mod;
    long long res = 1;
    for (int i = 30; i >= 0; --i) {
      res = (res * res) % mod;
      if ((b & (1 << i)) != 0) res = (res * et) % mod;
    }
    return res;
  }
  void panic() {
    cout << "impossible\n";
    exit(0);
  }
  int f[K][maxN];
  vector<int> g[maxN];
  int pr[maxN];
  int deep[maxN];
  void dfs(int v, int p = -1) {
    pr[v] = p;
    for (auto x : g[v]) {
      if (x == p) continue;
      deep[x] = deep[v] + 1;
      dfs(x, v);
      for (int k = 0; k < K; ++k) f[k][v] += max(1, f[k][x] - k);
    }
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      --a, --b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    dfs(0);
    vector<pair<int, int>> vi(n);
    vector<int> dp(n);
    for (int i = 0; i < n; ++i) {
      vi[i].second = i;
      vi[i].first = g[i].size();
      if (i) --vi[i].first;
      dp[vi[i].second] = vi[i].first;
    }
    sort(vi.begin(), vi.end());
    int q;
    cin >> q;
    vector<int> used(n);
    while (q--) {
      int v, k;
      cin >> v >> k;
      --v;
      if (k < K) {
        cout << f[k][v] << "\n";
        continue;
      }
      int t =
          lower_bound(vi.begin(), vi.end(), make_pair(k + 1, -1)) - vi.begin();
      vector<int> touch;
      priority_queue<pair<int, int>> q;
      for (int i = t; i < n; ++i) {
        q.push({deep[vi[i].second], vi[i].second});
        used[vi[i].second] = 1;
        touch.push_back(vi[i].second);
      }
      while (!q.empty()) {
        int v = q.top().second;
        q.pop();
        if (pr[v] != -1) {
          --dp[pr[v]];
          dp[pr[v]] += dp[v] - k;
          touch.push_back(pr[v]);
          if (!used[pr[v]] && dp[pr[v]] > k) {
            used[pr[v]] = 1;
            q.push({deep[pr[v]], pr[v]});
          }
        }
      }
      cout << dp[v] << "\n";
      for (auto x : touch) {
        used[x] = 0;
        if (x == 0)
          dp[x] = g[x].size();
        else
          dp[x] = g[x].size() - 1;
      }
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout.precision(20);
    srand(time(0));
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  