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. 
#include <bits/stdc++.h>
using namespace std;
const int K = 1000;
const int KK = 25;
vector<int> v[200005];
struct qry {
  int nod, k, pz, rz;
} qs[200005];
vector<int> qsk[5005];
int dpc[200005];
void dfsdpc(int nod, int t, int k) {
  dpc[nod] = 0;
  for (int i = 0; i < v[nod].size(); ++i) {
    if (v[nod][i] == t) continue;
    dfsdpc(v[nod][i], nod, k);
    dpc[nod] = dpc[nod] + max(1, dpc[v[nod][i]] - k);
  }
}
int dp[200005][KK + 2];
int aux[100];
void join(int a, int b) {
  int i, j;
  for (i = 0; i <= KK; ++i) {
    aux[i] = dp[a][i] + 1;
    for (j = 0; j < i; ++j) aux[i] = max(aux[i], dp[a][i - j - 1] + dp[b][j]);
  }
  for (i = 0; i <= KK; ++i) dp[a][i] = aux[i];
}
void dfs(int nod, int t) {
  for (int i = 0; i < v[nod].size(); ++i) {
    if (v[nod][i] == t) continue;
    dfs(v[nod][i], nod);
    join(nod, v[nod][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long m, q, t, s = 0, n, d, j, i, k;
  cin >> n;
  for (i = 1; i < n; ++i) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  cin >> q;
  for (i = 1; i <= q; ++i) {
    cin >> qs[i].nod >> qs[i].k;
    qs[i].pz = i;
    if (qs[i].k <= 200000 / K + 1) qsk[qs[i].k].push_back(i);
  }
  for (i = 0; i <= 200000 / K + 1; ++i) {
    if (qsk[i].size() == 0) continue;
    dfsdpc(1, -1, i);
    for (j = 0; j < qsk[i].size(); ++j)
      qs[qsk[i][j]].rz = dpc[qs[qsk[i][j]].nod];
  }
  dfs(1, -1);
  for (i = 1; i <= q; ++i) {
    if (qs[i].k <= 200000 / K + 1) {
      cout << qs[i].rz << '\n';
      continue;
    }
    for (int cat = 0; cat <= KK; ++cat)
      qs[i].rz = max(qs[i].rz, dp[qs[i].nod][cat] - cat * qs[i].k);
    cout << qs[i].rz << '\n';
  }
  return 0;
}
