#include <bits/stdc++.h>
using namespace std;
vector<int> L[200200];
int ord[2][200200];
int cnt = 0;
int par[200200];
int sum_kids[200200];
int dp[200200];
vector<pair<int, int> > queries[200200];
void dfs(int vx, int p = 0) {
  par[vx] = p;
  for (int i : L[vx])
    if (i != p) {
      dfs(i, vx);
      dp[vx] += dp[i];
    }
  ord[0][cnt++] = vx;
  dp[vx] = max(dp[vx], 1);
  sum_kids[p] += dp[vx];
}
int ans[200200];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    L[a].push_back(b);
    L[b].push_back(a);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    queries[k].push_back({v, i});
  }
  dfs(1);
  int sz = n;
  for (int k = 1; k < 200200; k++) {
    for (pair<int, int> i : queries[k - 1]) ans[i.second] = sum_kids[i.first];
    int nu = 0;
    for (int i = 0; i < sz; i++) {
      int vx = ord[(k + 1) % 2][i];
      sum_kids[par[vx]] -= dp[vx];
      dp[vx] = max(1, sum_kids[vx] - k);
      sum_kids[par[vx]] += dp[vx];
      if (dp[vx] > 1) ord[k % 2][nu++] = vx;
    }
    sz = nu;
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
}

### Prompt

In Cpp, your task is to solve 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;
vector<int> L[200200];
int ord[2][200200];
int cnt = 0;
int par[200200];
int sum_kids[200200];
int dp[200200];
vector<pair<int, int> > queries[200200];
void dfs(int vx, int p = 0) {
  par[vx] = p;
  for (int i : L[vx])
    if (i != p) {
      dfs(i, vx);
      dp[vx] += dp[i];
    }
  ord[0][cnt++] = vx;
  dp[vx] = max(dp[vx], 1);
  sum_kids[p] += dp[vx];
}
int ans[200200];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    L[a].push_back(b);
    L[b].push_back(a);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    queries[k].push_back({v, i});
  }
  dfs(1);
  int sz = n;
  for (int k = 1; k < 200200; k++) {
    for (pair<int, int> i : queries[k - 1]) ans[i.second] = sum_kids[i.first];
    int nu = 0;
    for (int i = 0; i < sz; i++) {
      int vx = ord[(k + 1) % 2][i];
      sum_kids[par[vx]] -= dp[vx];
      dp[vx] = max(1, sum_kids[vx] - k);
      sum_kids[par[vx]] += dp[vx];
      if (dp[vx] > 1) ord[k % 2][nu++] = vx;
    }
    sz = nu;
  }
  for (int i = 0; i < q; i++) printf("%d\n", ans[i]);
}

```