#include <bits/stdc++.h>
using namespace std;
struct frac {
  long long b, a;
  frac(long long b, long long a) : b(b), a(a) {}
  bool operator<(const frac &p) const { return b * p.a < p.b * a; }
};
struct fenwick {
  long long fen[200005];
  void update(int i, long long x) {
    while (i <= 200000) {
      fen[i] += x;
      i += i & -i;
    }
  }
  long long query(int i) const {
    long long r = 0;
    while (i) {
      r += fen[i];
      i -= i & -i;
    }
    return r;
  }
  long long query(int x, int y) const { return query(y) - query(x - 1); }
} fen0, fen1;
int n, q;
vector<int> edge[200005];
int st[200005], ed[200005];
int child[200005];
vector<pair<frac, int>> L;
void dfs(int x, int p, priority_queue<frac> &pq) {
  static int ord = 0;
  st[x] = ++ord;
  for (int i : edge[x]) {
    if (i == p) continue;
    ++child[x];
    priority_queue<frac> c;
    dfs(i, x, c);
    if (pq.size() < c.size()) swap(pq, c);
    while (!c.empty()) {
      pq.push(c.top());
      c.pop();
    }
  }
  frac f = frac(child[x] - 1, 1);
  while (!pq.empty() && f < pq.top()) {
    f.b += pq.top().b;
    f.a += pq.top().a;
    L.emplace_back(pq.top(), -p);
    pq.pop();
  }
  pq.push(f);
  L.emplace_back(f, p);
  ed[x] = ord;
}
long long ans[200005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  priority_queue<frac> c;
  dfs(1, 0, c);
  sort(L.begin(), L.end());
  cin >> q;
  vector<tuple<int, int, int>> Q;
  for (int i = 1; i <= q; ++i) {
    int v, k;
    cin >> v >> k;
    Q.emplace_back(k, v, i);
  }
  sort(Q.rbegin(), Q.rend());
  for (auto [k, v, i] : Q) {
    while (!L.empty() && frac(k, 1) < L.back().first) {
      auto [f, x] = L.back();
      L.pop_back();
      if (x < 0) {
        fen0.update(st[-x], -f.b);
        fen1.update(st[-x], -f.a);
      } else if (x > 0) {
        fen0.update(st[x], f.b);
        fen1.update(st[x], f.a);
      }
    }
    ans[i] = child[v] + fen0.query(st[v], ed[v]) - fen1.query(st[v], ed[v]) * k;
  }
  for (int i = 1; i <= q; ++i) {
    printf("%lld\n", ans[i]);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
struct frac {
  long long b, a;
  frac(long long b, long long a) : b(b), a(a) {}
  bool operator<(const frac &p) const { return b * p.a < p.b * a; }
};
struct fenwick {
  long long fen[200005];
  void update(int i, long long x) {
    while (i <= 200000) {
      fen[i] += x;
      i += i & -i;
    }
  }
  long long query(int i) const {
    long long r = 0;
    while (i) {
      r += fen[i];
      i -= i & -i;
    }
    return r;
  }
  long long query(int x, int y) const { return query(y) - query(x - 1); }
} fen0, fen1;
int n, q;
vector<int> edge[200005];
int st[200005], ed[200005];
int child[200005];
vector<pair<frac, int>> L;
void dfs(int x, int p, priority_queue<frac> &pq) {
  static int ord = 0;
  st[x] = ++ord;
  for (int i : edge[x]) {
    if (i == p) continue;
    ++child[x];
    priority_queue<frac> c;
    dfs(i, x, c);
    if (pq.size() < c.size()) swap(pq, c);
    while (!c.empty()) {
      pq.push(c.top());
      c.pop();
    }
  }
  frac f = frac(child[x] - 1, 1);
  while (!pq.empty() && f < pq.top()) {
    f.b += pq.top().b;
    f.a += pq.top().a;
    L.emplace_back(pq.top(), -p);
    pq.pop();
  }
  pq.push(f);
  L.emplace_back(f, p);
  ed[x] = ord;
}
long long ans[200005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  priority_queue<frac> c;
  dfs(1, 0, c);
  sort(L.begin(), L.end());
  cin >> q;
  vector<tuple<int, int, int>> Q;
  for (int i = 1; i <= q; ++i) {
    int v, k;
    cin >> v >> k;
    Q.emplace_back(k, v, i);
  }
  sort(Q.rbegin(), Q.rend());
  for (auto [k, v, i] : Q) {
    while (!L.empty() && frac(k, 1) < L.back().first) {
      auto [f, x] = L.back();
      L.pop_back();
      if (x < 0) {
        fen0.update(st[-x], -f.b);
        fen1.update(st[-x], -f.a);
      } else if (x > 0) {
        fen0.update(st[x], f.b);
        fen1.update(st[x], f.a);
      }
    }
    ans[i] = child[v] + fen0.query(st[v], ed[v]) - fen1.query(st[v], ed[v]) * k;
  }
  for (int i = 1; i <= q; ++i) {
    printf("%lld\n", ans[i]);
  }
  return 0;
}

```