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 MAXN = 2e5 + 10;
  int n, q, f[MAXN];
  int L[MAXN], R[MAXN], tot;
  long long sum[MAXN], sz[MAXN];
  int fa[MAXN], vis[MAXN];
  vector<int> e[MAXN];
  struct Qry {
    int v, k, idx;
    long long ans;
  } qry[MAXN];
  struct BIT {
    long long tree[MAXN];
    void update(int x, long long val) {
      for (; x <= n; x += (x & -x)) tree[x] += val;
    }
    long long query(int x) {
      long long ret = 0;
      for (; x; x -= (x & -x)) ret += tree[x];
      return ret;
    }
    long long query(int L, int R) { return query(R) - query(L - 1); }
  } t1, t2;
  struct Node {
    long long val;
    int idx;
    bool operator<(const Node& n2) const { return val < n2.val; }
  };
  priority_queue<Node> pq;
  bool cmp1(const Qry& q1, const Qry& q2) { return q1.k > q2.k; }
  bool cmp2(const Qry& q1, const Qry& q2) { return q1.idx < q2.idx; }
  int Find(int x) {
    if (fa[x] == x) return x;
    return fa[x] = Find(fa[x]);
  }
  void Merge(int x, int y) {
    if (Find(x) == Find(y)) return;
    fa[Find(x)] = Find(y);
  }
  long long D(long long x, long long y) {
    if (y < 0) x = -x, y = -y;
    return (x < 0) ? (-((-x + y - 1) / y)) : (x / y);
  }
  void dfs(int u) {
    L[u] = ++tot;
    for (auto v : e[u]) {
      if (v == f[u]) continue;
      f[v] = u;
      dfs(v);
      sum[u]++;
    }
    R[u] = tot;
  }
  void solve() {
    dfs(1);
    for (int i = (1); i <= (n); i++) {
      fa[i] = i;
      sz[i] = 1;
    }
    for (int i = (1); i <= (n); i++) {
      pq.push((Node){D(sum[i] - sz[i], sz[i]), i});
    }
    for (int i = (1); i <= (q); i++) {
      Qry& now = qry[i];
      while (pq.size()) {
        Node cur = pq.top();
        if (cur.val < now.k) break;
        pq.pop();
        if (vis[cur.idx]) continue;
        vis[cur.idx] = 1;
        if (cur.idx == 1) continue;
        Merge(cur.idx, f[cur.idx]);
        int top = Find(cur.idx);
        long long nsz = sz[cur.idx] + t1.query(L[cur.idx], R[cur.idx]);
        long long nsum = sum[cur.idx] + t2.query(L[cur.idx], R[cur.idx]);
        t1.update(L[f[cur.idx]], nsz);
        if (f[top]) t1.update(L[f[top]], -nsz);
        t2.update(L[f[cur.idx]], nsum);
        if (f[top]) t2.update(L[f[top]], -nsum);
        if (vis[top]) continue;
        nsz = sz[top] + t1.query(L[top], R[top]);
        nsum = sum[top] + t2.query(L[top], R[top]);
        pq.push((Node){D(nsum - nsz, nsz), top});
      }
      int u = now.v;
      long long nsz = sz[u] + t1.query(L[u], R[u]);
      long long nsum = sum[u] + t2.query(L[u], R[u]);
      now.ans = nsum - nsz * now.k + now.k - nsz + 1;
    }
  }
  int main() {
    scanf("%d", &n);
    for (int i = (1); i <= (n - 1); i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      e[u].push_back(v);
      e[v].push_back(u);
    }
    scanf("%d", &q);
    for (int i = (1); i <= (q); i++) {
      scanf("%d%d", &qry[i].v, &qry[i].k);
      qry[i].idx = i;
    }
    sort(qry + 1, qry + 1 + q, cmp1);
    solve();
    sort(qry + 1, qry + 1 + q, cmp2);
    for (int i = (1); i <= (q); i++) {
      printf("%lld\n", qry[i].ans);
    }
    return 0;
  }
  