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;
  template <typename T>
  bool chkmax(T &x, T y) {
    return x < y ? x = y, 1 : 0;
  }
  template <typename T>
  bool chkmin(T &x, T y) {
    return x > y ? x = y, 1 : 0;
  }
  int readint() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  struct node {
    int x, y;
    node(int x = 0, int y = 0) : x(x), y(y) {}
    bool operator<(const node c) const { return 1ll * y * c.x < 1ll * x * c.y; }
    bool operator>(const node c) const { return c < *this; }
    node operator+(const node c) const { return node(x + c.x, y + c.y); }
  } val[200005], sum[200005];
  mt19937 mrand(1);
  int n, q, tot, ncnt;
  int ch[200005][2], siz[200005], v[400005], nxt[400005], h[200005], son[200005],
      sz[200005], rt[200005], ans[200005];
  vector<pair<int, int> > qry[200005];
  void addedge(int x, int y) {
    v[++tot] = y;
    nxt[tot] = h[x];
    h[x] = tot;
    v[++tot] = x;
    nxt[tot] = h[y];
    h[y] = tot;
  }
  void dfs1(int u, int fa) {
    sz[u] = 1;
    for (int p = h[u]; p; p = nxt[p]) {
      if (v[p] == fa) continue;
      dfs1(v[p], u);
      sz[u] += sz[v[p]];
      if (sz[v[p]] > sz[son[u]]) son[u] = v[p];
    }
  }
  void update(int u) {
    siz[u] = 1 + siz[ch[u][0]] + siz[ch[u][1]];
    sum[u] = val[u] + sum[ch[u][0]] + sum[ch[u][1]];
  }
  void split(int u, node k, int &x, int &y) {
    if (!u) return (void)(x = y = 0);
    if (!(val[u] < k))
      x = u, split(ch[u][1], k, ch[x][1], y);
    else
      y = u, split(ch[u][0], k, x, ch[y][0]);
    update(u);
  }
  int merge(int x, int y) {
    if (!x || !y) return x + y;
    if (mrand() % (siz[x] + siz[y]) < siz[x]) {
      ch[x][1] = merge(ch[x][1], y);
      update(x);
      return x;
    } else {
      ch[y][0] = merge(x, ch[y][0]);
      update(y);
      return y;
    }
  }
  void dfs3(int u, int &t) {
    if (!u) return;
    int ta, tb;
    split(t, val[u], ta, tb);
    dfs3(ch[u][0], ta);
    dfs3(ch[u][1], tb);
    ch[u][0] = ch[u][1] = 0;
    update(u);
    t = merge(merge(ta, u), tb);
  }
  int getlf(int u) {
    while (ch[u][0]) u = ch[u][0];
    return u;
  }
  void dfs2(int u, int fa) {
    for (int p = h[u]; p; p = nxt[p])
      if (v[p] != fa && v[p] != son[u]) dfs2(v[p], u);
    int num = -1;
    if (son[u]) dfs2(son[u], u), rt[u] = rt[son[u]], num++;
    for (int p = h[u]; p; p = nxt[p]) {
      if (v[p] == fa || v[p] == son[u]) continue;
      dfs3(rt[v[p]], rt[u]);
      num++;
    }
    int a, b;
    for (auto r : qry[u]) {
      node tmp(1, r.first);
      split(rt[u], tmp, a, b);
      ans[r.second] = sum[a].y - 1ll * r.first * sum[a].x + num + 1;
      rt[u] = merge(a, b);
    }
    node th(1, num);
    while (siz[rt[u]] >= 1) {
      int tmp = getlf(rt[u]);
      if (th > val[tmp]) break;
      split(rt[u], val[tmp], a, b);
      th = th + val[a];
      rt[u] = merge(merge(ch[a][0], ch[a][1]), b);
    }
    ncnt++;
    val[ncnt] = sum[ncnt] = th, siz[ncnt] = 1;
    rt[u] = merge(ncnt, rt[u]);
  }
  int main() {
    n = readint();
    for (int i = 1; i < n; i++) addedge(readint(), readint());
    q = readint();
    for (int i = 1; i <= q; i++) {
      int x = readint(), y = readint();
      qry[x].push_back(make_pair(y, i));
    }
    dfs1(1, -1);
    dfs2(1, -1);
    for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
    return 0;
  }
  