#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using cmp = complex<int>;
const int N = 2e5 + 5;
int n, q, ans[N];
vector<int> G[N];
vector<pair<int, int>> Q[N];
vector<cmp> f[N];
int slp(const cmp& a) { return imag(a) / real(a); }
void dfs(int u, int fa) {
  for (int v : G[u])
    if (v != fa) {
      dfs(v, u);
      vector<cmp> t;
      swap(t, f[u]);
      for (int i = 0, j = 0; i < int((t).size()) || j < int((f[v]).size());)
        if (j == int((f[v]).size()) ||
            (i < int((t).size()) && slp(t[i]) <= slp(f[v][j])))
          f[u].push_back(t[i++]);
        else if (i && slp(f[u].back()) == slp(f[v][j]))
          f[u].back() += f[v][j++];
        else
          f[u].push_back(f[v][j++]);
    }
  for (auto [k, i] : Q[u]) {
    cmp sum;
    for (const cmp& x : f[u])
      if (slp(x) >= k) sum += x;
    ans[i] = imag(sum) + int((G[u]).size()) - (u > 1) - k * real(sum);
  }
  cmp t(1, int((G[u]).size()) - (u > 1) - 1);
  while (int((f[u]).size()) && slp(t) <= slp(f[u].back()))
    t += f[u].back(), f[u].pop_back();
  if (imag(t) > 0) f[u].push_back(t);
}
int main() {
  cin >> n;
  int u, v;
  for (int i = (2); i <= (n); i++)
    scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
  cin >> q;
  for (int i = (1); i <= (q); i++)
    scanf("%d%d", &u, &v), Q[u].emplace_back(v, i);
  dfs(1, 0);
  for (int i = (1); i <= (q); i++) printf("%d\n", ans[i]);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
using ll = long long;
using cmp = complex<int>;
const int N = 2e5 + 5;
int n, q, ans[N];
vector<int> G[N];
vector<pair<int, int>> Q[N];
vector<cmp> f[N];
int slp(const cmp& a) { return imag(a) / real(a); }
void dfs(int u, int fa) {
  for (int v : G[u])
    if (v != fa) {
      dfs(v, u);
      vector<cmp> t;
      swap(t, f[u]);
      for (int i = 0, j = 0; i < int((t).size()) || j < int((f[v]).size());)
        if (j == int((f[v]).size()) ||
            (i < int((t).size()) && slp(t[i]) <= slp(f[v][j])))
          f[u].push_back(t[i++]);
        else if (i && slp(f[u].back()) == slp(f[v][j]))
          f[u].back() += f[v][j++];
        else
          f[u].push_back(f[v][j++]);
    }
  for (auto [k, i] : Q[u]) {
    cmp sum;
    for (const cmp& x : f[u])
      if (slp(x) >= k) sum += x;
    ans[i] = imag(sum) + int((G[u]).size()) - (u > 1) - k * real(sum);
  }
  cmp t(1, int((G[u]).size()) - (u > 1) - 1);
  while (int((f[u]).size()) && slp(t) <= slp(f[u].back()))
    t += f[u].back(), f[u].pop_back();
  if (imag(t) > 0) f[u].push_back(t);
}
int main() {
  cin >> n;
  int u, v;
  for (int i = (2); i <= (n); i++)
    scanf("%d%d", &u, &v), G[u].push_back(v), G[v].push_back(u);
  cin >> q;
  for (int i = (1); i <= (q); i++)
    scanf("%d%d", &u, &v), Q[u].emplace_back(v, i);
  dfs(1, 0);
  for (int i = (1); i <= (q); i++) printf("%d\n", ans[i]);
}

```