#include <bits/stdc++.h>
std::priority_queue<std::pair<long long, int>,
                    std::vector<std::pair<long long, int>>,
                    std::greater<std::pair<long long, int>>>
    pq, del;
struct node {
  long long a, b;
  node(long long x = 0, long long y = 0) : a(y), b(x) {}
  node operator+(node const &y) const { return node(b + y.b, a + y.a); }
  node operator-(node const &y) const { return node(b - y.b, a - y.a); }
  node operator-() const { return node(-b, -a); }
  long long operator()(long long x) const { return a * x + b; }
  long long zero() const { return std::ceil((1. - b) / (a + 1.)); }
};
std::vector<int> v[200010];
node a[200010], c[200010];
int bg[200010], f[200010], ct, cnt, ed[200010], q[200010], n, m, fa[200010];
long long ans[1000010];
void upd(int i, node p) {
  if (i)
    for (; i <= n; i += i & -i) c[i] = c[i] + p;
}
node que(int i) {
  node ans;
  for (; i; i &= i - 1) ans = ans + c[i];
  return ans;
}
node query(int x) { return que(ed[x]) - que(bg[x] - 1); }
void update(int x, int y, node v) { upd(bg[fa[x]], -v), upd(bg[fa[y]], v); }
void dfs(int x, int f) {
  fa[x] = f;
  ::f[x] = x;
  bg[x] = ++cnt;
  for (auto u : v[x])
    if (u != f) ++a[x].b, dfs(u, x);
  ed[x] = cnt;
  upd(bg[fa[x]], node(1)), upd(bg[fa[fa[x]]], node(-1));
}
auto find = [](int x) {
  while (x != f[x]) x = f[x] = f[f[x]];
  return x;
};
auto clr = []() {
  while (del.size() && pq.top() == del.top()) pq.pop(), del.pop();
};
auto top = []() {
  clr();
  return pq.top();
};
auto pop = []() {
  clr();
  pq.pop();
};
int main() {
  scanf("%d", &n);
  for (int i = 2, x, y; i <= n; i++)
    scanf("%d%d", &x, &y), v[x].push_back(y), v[y].push_back(x);
  dfs(1, 0);
  for (int i = 2; i <= n; i++)
    if (a[i].b) pq.emplace(a[i].zero(), -i);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    long long b;
    scanf("%d%lld", q + i, &b);
    pq.emplace(-b, i);
  }
  while (ct < m) {
    auto [x, id] = top();
    pop();
    if (id < 0) {
      id = -id;
      if (id == 1) continue;
      int u = find(fa[id]);
      f[id] = u;
      if (u != 1) del.emplace(query(u).zero(), -u);
      update(u, id, query(id) + node(-1, 1));
      if (u != 1) pq.emplace(query(u).zero(), -u);
    } else {
      ans[id] = query(q[id])(x);
      ++ct;
    }
  }
  for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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. 

### Response

```cpp
#include <bits/stdc++.h>
std::priority_queue<std::pair<long long, int>,
                    std::vector<std::pair<long long, int>>,
                    std::greater<std::pair<long long, int>>>
    pq, del;
struct node {
  long long a, b;
  node(long long x = 0, long long y = 0) : a(y), b(x) {}
  node operator+(node const &y) const { return node(b + y.b, a + y.a); }
  node operator-(node const &y) const { return node(b - y.b, a - y.a); }
  node operator-() const { return node(-b, -a); }
  long long operator()(long long x) const { return a * x + b; }
  long long zero() const { return std::ceil((1. - b) / (a + 1.)); }
};
std::vector<int> v[200010];
node a[200010], c[200010];
int bg[200010], f[200010], ct, cnt, ed[200010], q[200010], n, m, fa[200010];
long long ans[1000010];
void upd(int i, node p) {
  if (i)
    for (; i <= n; i += i & -i) c[i] = c[i] + p;
}
node que(int i) {
  node ans;
  for (; i; i &= i - 1) ans = ans + c[i];
  return ans;
}
node query(int x) { return que(ed[x]) - que(bg[x] - 1); }
void update(int x, int y, node v) { upd(bg[fa[x]], -v), upd(bg[fa[y]], v); }
void dfs(int x, int f) {
  fa[x] = f;
  ::f[x] = x;
  bg[x] = ++cnt;
  for (auto u : v[x])
    if (u != f) ++a[x].b, dfs(u, x);
  ed[x] = cnt;
  upd(bg[fa[x]], node(1)), upd(bg[fa[fa[x]]], node(-1));
}
auto find = [](int x) {
  while (x != f[x]) x = f[x] = f[f[x]];
  return x;
};
auto clr = []() {
  while (del.size() && pq.top() == del.top()) pq.pop(), del.pop();
};
auto top = []() {
  clr();
  return pq.top();
};
auto pop = []() {
  clr();
  pq.pop();
};
int main() {
  scanf("%d", &n);
  for (int i = 2, x, y; i <= n; i++)
    scanf("%d%d", &x, &y), v[x].push_back(y), v[y].push_back(x);
  dfs(1, 0);
  for (int i = 2; i <= n; i++)
    if (a[i].b) pq.emplace(a[i].zero(), -i);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    long long b;
    scanf("%d%lld", q + i, &b);
    pq.emplace(-b, i);
  }
  while (ct < m) {
    auto [x, id] = top();
    pop();
    if (id < 0) {
      id = -id;
      if (id == 1) continue;
      int u = find(fa[id]);
      f[id] = u;
      if (u != 1) del.emplace(query(u).zero(), -u);
      update(u, id, query(id) + node(-1, 1));
      if (u != 1) pq.emplace(query(u).zero(), -u);
    } else {
      ans[id] = query(q[id])(x);
      ++ct;
    }
  }
  for (int i = 1; i <= m; i++) printf("%lld\n", ans[i]);
  return 0;
}

```