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 + 10;
  int n, m;
  vector<int> V[N];
  int fa[N], siz[N], er[N];
  int son[N];
  int ans[N];
  vector<pair<int, int> > q[N];
  void dfs1(int x) {
    siz[x] = 1;
    er[x] = 0;
    for (int i = 0; i < V[x].size(); ++i) {
      if (V[x][i] != fa[x]) {
        fa[V[x][i]] = x;
        dfs1(V[x][i]);
        siz[x] += siz[V[x][i]];
        ++er[x];
      }
    }
    return;
  }
  int lim[N];
  int F[23][N], cha[23][N];
  void dfs2(int x, int cs) {
    int y;
    son[x] = 0;
    for (int i = 0; i < V[x].size(); ++i) {
      y = V[x][i];
      if (y == fa[x]) continue;
      if (siz[y] > siz[son[x]]) {
        son[x] = y;
      }
    }
    if (!son[x]) {
      lim[x] = 0;
      F[cs][0] = 1;
      for (int i = 0; i < q[x].size(); ++i) {
        ans[q[x][i].second] = 0;
      }
      return;
    }
    dfs2(son[x], cs);
    int L = lim[son[x]];
    cha[cs][lim[son[x]] + 1]++;
    for (int i = 0; i < V[x].size(); ++i) {
      y = V[x][i];
      if (y == fa[x] || y == son[x]) continue;
      dfs2(y, cs + 1);
      while (L < lim[y]) {
        ++L;
        F[cs][L] = 0;
      }
      for (int j = 0; j <= lim[y]; ++j) {
        F[cs][j] += F[cs + 1][j];
      }
      cha[cs][lim[y] + 1]++;
    }
    int hhh = 0;
    for (int j = 0; j <= L; ++j) {
      hhh += cha[cs][j];
      F[cs][j] += hhh;
    }
    for (int i = 0; i < q[x].size(); ++i) {
      if (q[x][i].first > L) {
        ans[q[x][i].second] = er[x];
      } else {
        ans[q[x][i].second] = F[cs][q[x][i].first];
      }
    }
    if (er[x] > L) {
      lim[x] = er[x];
      for (int j = L + 1; j <= er[x]; ++j) {
        F[cs][j] = er[x];
      }
    } else {
      lim[x] = L;
    }
    while (lim[x] > 0 && F[cs][lim[x]] - lim[x] < 1) {
      lim[x]--;
    }
    for (int j = 0; j <= lim[x]; ++j) {
      F[cs][j] -= j;
      if (F[cs][j] < 1) F[cs][j] = 1;
    }
    cha[cs][lim[son[x]] + 1]--;
    for (int i = 0; i < V[x].size(); ++i) {
      y = V[x][i];
      if (y == fa[x] || y == son[x]) continue;
      cha[cs][lim[y] + 1]--;
    }
    return;
  }
  int main() {
    int u, v;
    scanf("%d", &n);
    for (int i = 1; i < n; ++i) {
      scanf("%d%d", &u, &v);
      V[u].push_back(v);
      V[v].push_back(u);
    }
    dfs1(1);
    scanf("%d", &m);
    for (int i = 1; i <= m; ++i) {
      scanf("%d%d", &u, &v);
      q[u].push_back(make_pair(v, i));
    }
    dfs2(1, 0);
    for (int i = 1; i <= m; ++i) {
      printf("%d\n", ans[i]);
    }
    return 0;
  }
  