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;
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = 0;
      ch = getchar();
    }
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f ? x : -x;
  }
  const int N = 200005;
  vector<int> adj[N];
  int n, m, fa[N], a[N], siz[N], dfn[N], tmr;
  long long ans[N];
  vector<int> e[N];
  struct node {
    long long val, key;
    int siz, u;
    node() { val = siz = 0; }
    node(long long v_, int s_, int u_) {
      val = v_, siz = s_, u = u_, key = ceil((double)-val / siz);
    }
    inline bool operator==(const node &t) const {
      return key == t.key && u == t.u;
    }
    inline bool operator<(const node &t) const {
      return key != t.key ? key > t.key : u > t.u;
    }
  };
  struct QUE {
    int u, x, id;
    inline bool operator<(const QUE &t) const { return x < t.x; }
  } q[N];
  namespace Set {
  priority_queue<node> pq, del;
  inline void insert(const node &t) { pq.push(t); }
  inline void erase(const node &t) { del.push(t); }
  inline node top() {
    while (!del.empty() && del.top() == pq.top()) del.pop(), pq.pop();
    return pq.top();
  }
  inline bool empty() { return pq.size() == del.size(); }
  }  // namespace Set
  int F[N];
  struct BIT {
    long long tr[N];
    inline void add(int x, long long d) {
      if (!x) return;
      for (int i = x; i <= n; i += i & -i) tr[i] += d;
    }
    inline long long ask(int x) {
      long long res = 0;
      for (int i = x; i > 0; i -= i & -i) res += tr[i];
      return res;
    }
    inline long long ask(int l, int r) { return ask(r) - ask(l - 1); }
  } tr[2];
  inline int anc(int x) { return x == F[x] ? x : F[x] = anc(F[x]); }
  inline void init_DSU() {
    for (int i = 1, iend = n; i <= iend; ++i) F[i] = i;
  }
  int heng[N], hengid;
  void dfs1(int u, int fa) {
    ++hengid, heng[hengid] = u;
    siz[u] = 1, dfn[u] = ++tmr;
    for (int v : adj[u]) {
      if (v == fa) continue;
      a[u]++;
      ::fa[v] = u;
      dfs1(v, u), siz[u] += siz[v];
    }
    if (u != 1) Set::insert(node(a[u], 1, u));
  }
  inline void link(int u) {
    int sz = (tr[0].ask(dfn[u], dfn[u] + siz[u] - 1));
    long long sm = (tr[1].ask(dfn[u], dfn[u] + siz[u] - 1));
    F[u] = anc(fa[u]);
    if (F[u] != 1) {
      Set::erase(node((tr[1].ask(dfn[F[u]], dfn[F[u]] + siz[F[u]] - 1)),
                      (tr[0].ask(dfn[F[u]], dfn[F[u]] + siz[F[u]] - 1)), F[u]));
    }
    tr[0].add(dfn[fa[u]], sz), tr[1].add(dfn[fa[u]], sm);
    if (F[u] != 1) {
      int gfa = fa[F[u]];
      tr[0].add(dfn[gfa], -sz), tr[1].add(dfn[gfa], -sm);
      Set::insert(node((tr[1].ask(dfn[F[u]], dfn[F[u]] + siz[F[u]] - 1)),
                       (tr[0].ask(dfn[F[u]], dfn[F[u]] + siz[F[u]] - 1)), F[u]));
    }
  }
  signed main() {
    n = read(), init_DSU();
    for (int i = 1, iend = n - 1; i <= iend; ++i) {
      int u = read(), v = read();
      adj[u].push_back(v), adj[v].push_back(u);
    }
    m = read();
    for (int i = 1, iend = m; i <= iend; ++i) {
      q[i].u = read(), q[i].x = -read() - 1, q[i].id = i;
      ans[i] -= q[i].x;
    }
    dfs1(1, 0), sort(q + 1, q + m + 1);
    for (int _ = 1, _end = n; _ <= _end; ++_) {
      int i = heng[_];
      tr[0].add(dfn[i], 1), tr[1].add(dfn[i], a[i]);
      if (i != 1) tr[0].add(dfn[fa[i]], -1), tr[1].add(dfn[fa[i]], -a[i]);
    }
    for (int i = 1, iend = m; i <= iend; ++i) {
      int u = q[i].u, x = q[i].x;
      while (!Set::empty()) {
        node now = Set::top();
        if (now.key <= x)
          Set::erase(now), link(now.u);
        else
          break;
      }
      int l = dfn[u], r = dfn[u] + siz[u] - 1;
      ans[q[i].id] += tr[0].ask(l, r) * q[i].x + tr[1].ask(l, r);
    }
    for (int i = 1, iend = m; i <= iend; ++i) printf("%lld\n", ans[i]);
    return 0;
  }
  