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 = 200000;
  int PRT;
  struct Tag {
    long long b;
    int k;
    Tag() { b = k = 0; }
    Tag(long long x, int y) { b = x, k = y; }
    bool empty() { return k == 0 && b == 0; }
    Tag operator+(const Tag &A) const { return Tag(b + A.b, k + A.k); }
    Tag operator>>(const int &x) const { return Tag(b - 1ll * x * k, k); }
  };
  struct Node {
    int sz, val;
    Tag tag;
    Node *lson, *rson;
    Node() {}
    Node(int x) { tag = Tag(), val = 0, sz = x, lson = rson = NULL; }
    void pushup() { val = (lson ? lson->val : 0); }
    void addtag(Tag t) { val += t.b, tag = tag + t; }
    void pushdown() {
      if (!lson) lson = new Node(sz + 1 >> 1);
      if (!rson) rson = new Node(sz >> 1);
      if (tag.empty()) return;
      lson->addtag(tag), rson->addtag(tag >> (sz + 1 >> 1));
      tag = Tag();
    }
  };
  long long Query(Node *T, int l, int r, int x) {
    if (!T) return 0;
    if (l == r) return T->val;
    T->pushdown();
    int mid = l + r >> 1;
    return x <= mid ? Query(T->lson, l, mid, x) : Query(T->rson, mid + 1, r, x);
  }
  Node *Merge(Node *L, Node *R) {
    if (!L || !R) return L ? L : R;
    L->lson = Merge(L->lson, R->lson);
    L->rson = Merge(L->rson, R->rson);
    L->tag = L->tag + R->tag, L->val += R->val;
    return L;
  }
  void Flush(Node *&T) {
    if (!T) return;
    if (T->val < 0) {
      T = NULL;
      return;
    }
    if (T->sz == 1) return;
    T->pushdown();
    if (T->rson->val >= 0)
      Flush(T->rson);
    else
      Flush(T->lson), Flush(T->rson);
    T->pushup();
  }
  int n, q;
  vector<int> G[200010];
  vector<pair<int, int> > qry[200010];
  int ans[200010];
  int a[200010];
  Node *Solve(int x, int last) {
    Node *tmp = NULL;
    for (auto v : G[x])
      if (v != last) {
        Node *qwq = Solve(v, x);
        if (!qwq) qwq = new Node(N + 1);
        if (v == 7) PRT = 1;
        qwq->addtag(Tag(0, 1));
        qwq->addtag(Tag(-1, 0));
        if (v == 7) PRT = 0;
        Flush(qwq);
        if (!qwq) qwq = new Node(N + 1);
        qwq->addtag(Tag(1, 0));
        tmp = Merge(tmp, qwq);
      }
    for (auto it : qry[x]) {
      int k = it.first, id = it.second;
      ans[id] = Query(tmp, 0, N, k);
    }
    return tmp;
  }
  int main() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      G[x].push_back(y);
      G[y].push_back(x);
    }
    int q;
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
      int v, k;
      scanf("%d%d", &v, &k);
      qry[v].push_back(make_pair(k, i));
    }
    Solve(1, 1);
    for (int i = 0; i < q; i++) {
      printf("%d\n", ans[i]);
    }
    return 0;
  }
  