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>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
using ll = long long;
using ld = long double;
const int N = 200000;
const int K = 340;
const int C = N / K + 2;
int dpz[N + 5][C + 5];
int sz[N + 5];
int dp[N + 5];
vector<int> gg[N + 5];
vector<int> graf[N + 5];
const int INF = 100000000;
int dpn[C + 5];
void solveBig(int v) {
  int tsz = 0;
  dpz[v][0] = graf[v].size();
  for (int i = 1; i <= C; i++) dpz[v][i] = -INF;
  for (auto c : graf[v]) {
    solveBig(c);
    for (int i = 0; i <= C; i++) dpn[i] = dpz[v][i];
    for (int i = 0; i <= min(C, sz[c]); i++) {
      for (int j = 0; j <= tsz; j++) {
        if (i + j + 1 > C) continue;
        dpz[v][i + j + 1] = max(dpz[v][i + j + 1], dpn[j] + dpz[c][i] - 1);
      }
    }
    tsz += sz[c];
    tsz = min(tsz, C);
  }
}
int _k;
void solve(int v) {
  dp[v] = 0;
  if (graf[v].size() == 0) return;
  for (auto c : graf[v]) {
    solve(c);
    dp[v] += max(1, dp[c] - _k);
  }
}
void find_tree(int v, int p) {
  sz[v] = 1;
  for (auto c : gg[v]) {
    if (c == p) continue;
    sz[v]++;
    graf[v].push_back(c);
    find_tree(c, v);
  }
}
int qa[N + 5], qb[N + 5], soln[N + 5];
vector<pair<int, int>> vec[N + 5];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cout.precision(10);
  cout << fixed;
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    gg[a].push_back(b);
    gg[b].push_back(a);
  }
  find_tree(1, 0);
  int qrs;
  cin >> qrs;
  solveBig(1);
  for (int i = 1; i <= qrs; i++) {
    cin >> qa[i] >> qb[i];
    if (qb[i] <= K) {
      vec[qb[i]].push_back({qa[i], i});
    } else {
      ll res = 0;
      for (int j = 0; j <= min(sz[i], C); j++)
        res = max(res, dpz[qa[i]][j] - 1LL * j * qb[i]);
      soln[i] = res;
    }
  }
  for (int i = 0; i <= K; i++) {
    _k = i;
    solve(1);
    for (auto c : vec[i]) soln[c.second] = dp[c.first];
  }
  for (int i = 1; i <= qrs; i++) cout << soln[i] << "\n";
  return 0;
}
