#include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int>> E;
vector<int> p;
vector<int> d;
vector<int> c;
void dfs(int cur, int par, int depth) {
  d[cur] = depth;
  p[cur] = par;
  for (int i = 0; i < (E[cur].size()); ++i) {
    int to = E[cur][i];
    if (to == par) continue;
    c[cur]++;
    dfs(to, cur, depth + 1);
  }
}
int main() {
  cin >> n;
  E.resize(n);
  p.resize(n, -1);
  for (int i = 0; i < (n - 1); ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--;
    y--;
    E[x].push_back(y);
    E[y].push_back(x);
  }
  d.resize(n, 0);
  c.resize(n, 0);
  dfs(0, -1, 0);
  vector<pair<int, int>> L(n);
  for (int i = 0; i < (n); ++i) {
    L[i] = make_pair(c[i], i);
  }
  sort(L.rbegin(), L.rend());
  int q;
  cin >> q;
  vector<vector<pair<int, int>>> qs(200003);
  for (int i = 0; i < (q); ++i) {
    int v, k;
    scanf("%d %d", &v, &k);
    v--;
    qs[k].emplace_back(v, i);
  }
  vector<int> ans(q);
  vector<int> dp(n, 0);
  vector<bool> f(n, false);
  for (int i = 0; i < (200003); ++i) {
    priority_queue<pair<int, int>> Q;
    for (int j = 0; j < (n); ++j) {
      if (L[j].first <= i + 1) break;
      Q.emplace(d[L[j].second], L[j].second);
      f[L[j].second] = true;
    }
    vector<int> x;
    while (Q.size() > 0) {
      pair<int, int> t = Q.top();
      Q.pop();
      dp[t.second] -= i + 1;
      dp[t.second] += c[t.second];
      x.push_back(t.second);
      if (dp[t.second] <= 0) {
        continue;
      }
      x.push_back(t.second);
      int pp = p[t.second];
      if (pp == -1) continue;
      dp[pp] += dp[t.second];
      if (f[pp]) continue;
      f[pp] = true;
      Q.emplace(d[pp], pp);
    }
    for (int j = 0; j < (qs[i].size()); ++j) {
      int v = qs[i][j].first;
      int ind = qs[i][j].second;
      if (f[v] == false) {
        ans[ind] = c[v] - (i + 1);
      } else {
        ans[ind] = max(c[v] - (i + 1), dp[v]);
      }
      ans[ind] += i + 1;
    }
    for (int j = 0; j < (x.size()); ++j) {
      f[x[j]] = false;
      dp[x[j]] = 0;
    }
  }
  for (int i = 0; i < (q); ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
int n;
vector<vector<int>> E;
vector<int> p;
vector<int> d;
vector<int> c;
void dfs(int cur, int par, int depth) {
  d[cur] = depth;
  p[cur] = par;
  for (int i = 0; i < (E[cur].size()); ++i) {
    int to = E[cur][i];
    if (to == par) continue;
    c[cur]++;
    dfs(to, cur, depth + 1);
  }
}
int main() {
  cin >> n;
  E.resize(n);
  p.resize(n, -1);
  for (int i = 0; i < (n - 1); ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--;
    y--;
    E[x].push_back(y);
    E[y].push_back(x);
  }
  d.resize(n, 0);
  c.resize(n, 0);
  dfs(0, -1, 0);
  vector<pair<int, int>> L(n);
  for (int i = 0; i < (n); ++i) {
    L[i] = make_pair(c[i], i);
  }
  sort(L.rbegin(), L.rend());
  int q;
  cin >> q;
  vector<vector<pair<int, int>>> qs(200003);
  for (int i = 0; i < (q); ++i) {
    int v, k;
    scanf("%d %d", &v, &k);
    v--;
    qs[k].emplace_back(v, i);
  }
  vector<int> ans(q);
  vector<int> dp(n, 0);
  vector<bool> f(n, false);
  for (int i = 0; i < (200003); ++i) {
    priority_queue<pair<int, int>> Q;
    for (int j = 0; j < (n); ++j) {
      if (L[j].first <= i + 1) break;
      Q.emplace(d[L[j].second], L[j].second);
      f[L[j].second] = true;
    }
    vector<int> x;
    while (Q.size() > 0) {
      pair<int, int> t = Q.top();
      Q.pop();
      dp[t.second] -= i + 1;
      dp[t.second] += c[t.second];
      x.push_back(t.second);
      if (dp[t.second] <= 0) {
        continue;
      }
      x.push_back(t.second);
      int pp = p[t.second];
      if (pp == -1) continue;
      dp[pp] += dp[t.second];
      if (f[pp]) continue;
      f[pp] = true;
      Q.emplace(d[pp], pp);
    }
    for (int j = 0; j < (qs[i].size()); ++j) {
      int v = qs[i][j].first;
      int ind = qs[i][j].second;
      if (f[v] == false) {
        ans[ind] = c[v] - (i + 1);
      } else {
        ans[ind] = max(c[v] - (i + 1), dp[v]);
      }
      ans[ind] += i + 1;
    }
    for (int j = 0; j < (x.size()); ++j) {
      f[x[j]] = false;
      dp[x[j]] = 0;
    }
  }
  for (int i = 0; i < (q); ++i) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```