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 = 2e5 + 5;
  const int INF = 0x3F3F3F3F;
  vector<int> G[N];
  int n, m, cnt;
  int a[N], sz[N], fa[N], dfn[N];
  long long ans[N];
  bool del[N];
  set<pair<int, int> > S;
  struct rec {
    int x, k, id;
    bool operator<(const rec &b) const { return k < b.k; }
  } q[N];
  namespace FastIn {
  char Buf[1 << 22], *S(Buf), *T(Buf);
  inline int ReadInt() {
    int x = 0, sign = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') {
        sign = -1;
      }
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * sign;
  }
  }  // namespace FastIn
  void DFS(int x, int pre) {
    fa[x] = pre;
    dfn[x] = ++cnt;
    sz[x] = 1, a[x] = -1;
    for (auto y : G[x]) {
      if (y != pre) {
        ++a[x], DFS(y, x);
        sz[x] += sz[y];
      }
    }
  }
  struct DSU {
    int pre[N];
    void Init(int sz) {
      for (int i = 1; i <= sz; ++i) {
        pre[i] = i;
      }
    }
    inline int find(int x) {
      if (x == pre[x]) {
        return x;
      } else {
        return pre[x] = find(pre[x]);
      }
    }
    inline void merge(int x, int y) { pre[find(x)] = find(y); }
  } DS;
  struct BIT {
    long long dt[N];
    inline void update(int x, long long v) {
      for (; x <= n; x += x & -x) {
        dt[x] += v;
      }
    }
    inline long long query(int x) {
      long long ret = 0;
      for (; x; x -= x & -x) {
        ret += dt[x];
      }
      return ret;
    }
  } T[2];
  void GetTop(long long &v, long long &id) {
    if (S.empty()) {
      id = v = INF;
      return;
    }
    int x = S.begin()->second;
    while (del[x]) {
      S.erase(S.begin());
      if (S.empty()) {
        id = v = INF;
        return;
      }
      x = S.begin()->second;
    }
    id = x, v = S.begin()->first;
  }
  inline long long Get(int x, bool tp) {
    return T[tp].query(dfn[x] + sz[x] - 1) - T[tp].query(dfn[x] - 1);
  }
  inline void update(int x) {
    S.insert(make_pair(ceil((long double)-Get(x, 0) / Get(x, 1)), x));
  }
  inline long long query(int x, int k) { return Get(x, 0) + Get(x, 1) * k; }
  int main() {
    n = FastIn ::ReadInt();
    for (int i = 1; i < n; ++i) {
      int x = FastIn ::ReadInt();
      int y = FastIn ::ReadInt();
      G[x].push_back(y);
      G[y].push_back(x);
    }
    DS.Init(n), DFS(1, 0);
    m = FastIn ::ReadInt();
    for (int i = 1; i <= m; ++i) {
      q[i].x = FastIn ::ReadInt();
      q[i].k = -FastIn ::ReadInt();
      q[i].id = i;
    }
    sort(q + 1, q + m + 1);
    for (int i = 1; i <= n; ++i) {
      T[0].update(dfn[i], a[i]);
      T[1].update(dfn[i], 1);
      if (fa[i]) {
        T[0].update(dfn[fa[i]], -a[i]);
        T[1].update(dfn[fa[i]], -1);
      }
      if (i > 1) {
        S.insert(make_pair(-a[i], i));
      }
    }
    for (int i = 1; i <= m; ++i) {
      long long v, x, sz, dt, nxt;
      GetTop(v, x);
      while (v <= q[i].k) {
        sz = Get(x, 1);
        dt = Get(x, 0);
        nxt = DS.find(fa[x]);
        DS.merge(x, fa[x]);
        S.erase(S.begin());
        del[x] = 1;
        T[0].update(dfn[fa[x]], dt);
        T[1].update(dfn[fa[x]], sz);
        if (fa[nxt]) {
          T[0].update(dfn[fa[nxt]], -dt);
          T[1].update(dfn[fa[nxt]], -sz);
          update(nxt);
        }
        GetTop(v, x);
      }
      ans[q[i].id] =
          max(query(q[i].x, q[i].k) - q[i].k + 1, (long long)a[q[i].x] + 1);
    }
    for (int i = 1; i <= m; ++i) {
      printf("%lld\n", ans[i]);
    }
    return 0;
  }
  