#include <bits/stdc++.h>
using namespace std;
const int N = 200000;
int PRT;
struct Tag {
  long long b;
  int k;
  Tag() { b = k = 0; }
  Tag(long long x, int y) { b = x, k = y; }
  bool empty() { return k == 0 && b == 0; }
  Tag operator+(const Tag &A) const { return Tag(b + A.b, k + A.k); }
  Tag operator>>(const int &x) const { return Tag(b - 1ll * x * k, k); }
};
struct Node {
  int sz, val;
  Tag tag;
  Node *lson, *rson;
  Node() {}
  Node(int x) { tag = Tag(), val = 0, sz = x, lson = rson = NULL; }
  void pushup() { val = (lson ? lson->val : 0); }
  void addtag(Tag t) { val += t.b, tag = tag + t; }
  void pushdown() {
    if (!lson) lson = new Node(sz + 1 >> 1);
    if (!rson) rson = new Node(sz >> 1);
    if (tag.empty()) return;
    lson->addtag(tag), rson->addtag(tag >> (sz + 1 >> 1));
    tag = Tag();
  }
};
long long Query(Node *T, int l, int r, int x) {
  if (!T) return 0;
  if (l == r) return T->val;
  T->pushdown();
  int mid = l + r >> 1;
  return x <= mid ? Query(T->lson, l, mid, x) : Query(T->rson, mid + 1, r, x);
}
Node *Merge(Node *L, Node *R) {
  if (!L || !R) return L ? L : R;
  L->lson = Merge(L->lson, R->lson);
  L->rson = Merge(L->rson, R->rson);
  L->tag = L->tag + R->tag, L->val += R->val;
  return L;
}
void Flush(Node *&T) {
  if (!T) return;
  if (T->val < 0) {
    T = NULL;
    return;
  }
  if (T->sz == 1) return;
  T->pushdown();
  if (T->rson->val >= 0)
    Flush(T->rson);
  else
    Flush(T->lson), Flush(T->rson);
  T->pushup();
}
int n, q;
vector<int> G[200010];
vector<pair<int, int> > qry[200010];
int ans[200010];
int a[200010];
Node *Solve(int x, int last) {
  Node *tmp = NULL;
  for (auto v : G[x])
    if (v != last) {
      Node *qwq = Solve(v, x);
      if (!qwq) qwq = new Node(N + 1);
      if (v == 7) PRT = 1;
      qwq->addtag(Tag(0, 1));
      qwq->addtag(Tag(-1, 0));
      if (v == 7) PRT = 0;
      Flush(qwq);
      if (!qwq) qwq = new Node(N + 1);
      qwq->addtag(Tag(1, 0));
      tmp = Merge(tmp, qwq);
    }
  for (auto it : qry[x]) {
    int k = it.first, id = it.second;
    ans[id] = Query(tmp, 0, N, k);
  }
  return tmp;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    qry[v].push_back(make_pair(k, i));
  }
  Solve(1, 1);
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### 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;
const int N = 200000;
int PRT;
struct Tag {
  long long b;
  int k;
  Tag() { b = k = 0; }
  Tag(long long x, int y) { b = x, k = y; }
  bool empty() { return k == 0 && b == 0; }
  Tag operator+(const Tag &A) const { return Tag(b + A.b, k + A.k); }
  Tag operator>>(const int &x) const { return Tag(b - 1ll * x * k, k); }
};
struct Node {
  int sz, val;
  Tag tag;
  Node *lson, *rson;
  Node() {}
  Node(int x) { tag = Tag(), val = 0, sz = x, lson = rson = NULL; }
  void pushup() { val = (lson ? lson->val : 0); }
  void addtag(Tag t) { val += t.b, tag = tag + t; }
  void pushdown() {
    if (!lson) lson = new Node(sz + 1 >> 1);
    if (!rson) rson = new Node(sz >> 1);
    if (tag.empty()) return;
    lson->addtag(tag), rson->addtag(tag >> (sz + 1 >> 1));
    tag = Tag();
  }
};
long long Query(Node *T, int l, int r, int x) {
  if (!T) return 0;
  if (l == r) return T->val;
  T->pushdown();
  int mid = l + r >> 1;
  return x <= mid ? Query(T->lson, l, mid, x) : Query(T->rson, mid + 1, r, x);
}
Node *Merge(Node *L, Node *R) {
  if (!L || !R) return L ? L : R;
  L->lson = Merge(L->lson, R->lson);
  L->rson = Merge(L->rson, R->rson);
  L->tag = L->tag + R->tag, L->val += R->val;
  return L;
}
void Flush(Node *&T) {
  if (!T) return;
  if (T->val < 0) {
    T = NULL;
    return;
  }
  if (T->sz == 1) return;
  T->pushdown();
  if (T->rson->val >= 0)
    Flush(T->rson);
  else
    Flush(T->lson), Flush(T->rson);
  T->pushup();
}
int n, q;
vector<int> G[200010];
vector<pair<int, int> > qry[200010];
int ans[200010];
int a[200010];
Node *Solve(int x, int last) {
  Node *tmp = NULL;
  for (auto v : G[x])
    if (v != last) {
      Node *qwq = Solve(v, x);
      if (!qwq) qwq = new Node(N + 1);
      if (v == 7) PRT = 1;
      qwq->addtag(Tag(0, 1));
      qwq->addtag(Tag(-1, 0));
      if (v == 7) PRT = 0;
      Flush(qwq);
      if (!qwq) qwq = new Node(N + 1);
      qwq->addtag(Tag(1, 0));
      tmp = Merge(tmp, qwq);
    }
  for (auto it : qry[x]) {
    int k = it.first, id = it.second;
    ans[id] = Query(tmp, 0, N, k);
  }
  return tmp;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    qry[v].push_back(make_pair(k, i));
  }
  Solve(1, 1);
  for (int i = 0; i < q; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```