#include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int>> g;
vector<vector<int>> dp;
int q(int v, int k) { return dp[v][min(k, (int)dp[v].size() - 1)]; }
void dfs(int v, int p) {
  int cnt = 0;
  vector<int> all1;
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    all1.push_back(u);
    ++cnt;
    dfs(u, v);
  }
  for (int i = 0; i <= n; i++) {
    int here = cnt - all1.size();
    vector<int> all2;
    for (int u : all1) {
      int val = q(u, i);
      if (val - i > 1) {
        here += val - i;
        all2.push_back(u);
      } else {
        here += 1;
      }
    }
    all1 = all2;
    dp[v].push_back(here);
    if (here == cnt) {
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  g.resize(n);
  dp.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    --v;
    --u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, -1);
  int q1;
  cin >> q1;
  for (int i = 0; i < q1; i++) {
    int v, k;
    cin >> v >> k;
    cout << q(v - 1, k) << '\n';
  }
}

### Prompt

Please formulate a cpp solution to the following problem:

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;
int n;
vector<vector<int>> g;
vector<vector<int>> dp;
int q(int v, int k) { return dp[v][min(k, (int)dp[v].size() - 1)]; }
void dfs(int v, int p) {
  int cnt = 0;
  vector<int> all1;
  for (int u : g[v]) {
    if (u == p) {
      continue;
    }
    all1.push_back(u);
    ++cnt;
    dfs(u, v);
  }
  for (int i = 0; i <= n; i++) {
    int here = cnt - all1.size();
    vector<int> all2;
    for (int u : all1) {
      int val = q(u, i);
      if (val - i > 1) {
        here += val - i;
        all2.push_back(u);
      } else {
        here += 1;
      }
    }
    all1 = all2;
    dp[v].push_back(here);
    if (here == cnt) {
      break;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  g.resize(n);
  dp.resize(n);
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    --v;
    --u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, -1);
  int q1;
  cin >> q1;
  for (int i = 0; i < q1; i++) {
    int v, k;
    cin >> v >> k;
    cout << q(v - 1, k) << '\n';
  }
}

```