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")
using namespace std;
int n, q, par[200005];
int dp[200005][500 + 5];
vector<int> muchii[200005];
vector<pair<int, int> > query[200005];
pair<int, int> qr[200005];
int sol[200005];
void dfs(int nod) {
  for (auto i : muchii[nod])
    if (i != par[nod]) {
      par[i] = nod;
      dfs(i);
    }
}
void solve(int v) {
  for (int k = 0; k <= 500; k++) dp[v][k] = 0;
  for (int j = 0; j < muchii[v].size(); j++) {
    int i = muchii[v][j];
    if (i != par[v]) {
      solve(i);
      for (int k = 0; k <= 500; k++) {
        int val = max(dp[i][k] - k, 1);
        dp[v][k] += val;
      }
    }
  }
  for (int j = 0; j < query[v].size(); j++) {
    int k = query[v][j].first;
    int poz = query[v][j].second;
    if (k <= 500) sol[poz] = dp[v][k];
  }
}
int sub[200005];
int prv[500 + 5], cur[500 + 5];
void getbest(int nod) {
  for (int i = 0; i <= 500; i++) dp[nod][i] = 0;
  dp[nod][0] = (int)muchii[nod].size() - (nod != 1);
  for (int i = 0; i <= 500; i++) prv[i] = 0;
  prv[0] = 0;
  int stop = min(n, 500);
  for (int j = 0; j < muchii[nod].size(); j++) {
    int i = muchii[nod][j];
    if (i == par[nod]) continue;
    getbest(i);
    for (int elim = 0; elim <= min(stop, sub[i] + sub[nod]); elim++)
      cur[elim] = 0;
    for (int bef = 0; bef <= min(stop, sub[nod]); bef++) {
      for (int fb = 0; fb <= sub[i] && bef + fb <= stop; fb++) {
        int elim = bef + fb;
        if (elim == 0) continue;
        int val = prv[elim - fb];
        if (fb == 0)
          val++;
        else
          val += dp[i][fb - 1];
        cur[elim] = max(cur[elim], val);
        if (elim != 0) dp[nod][elim] = max(dp[nod][elim], val);
      }
    }
    for (int elim = 0; elim <= min(stop, sub[i] + sub[nod]); elim++)
      prv[elim] = dp[nod][elim];
    sub[nod] += sub[i];
  }
  sub[nod]++;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    muchii[a].push_back(b);
    muchii[b].push_back(a);
  }
  dfs(1);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int v, k;
    cin >> v >> k;
    query[v].push_back({k, i});
    qr[i] = {v, k};
  }
  solve(1);
  getbest(1);
  int tc = 0;
  for (int z = 1; z <= q; z++) {
    tc++;
    long long v, k;
    v = qr[tc].first;
    k = qr[tc].second;
    if (k <= 500)
      continue;
    else {
      int ans = (int)muchii[v].size() - (v != 1);
      for (int elim = 0; elim <= 500; elim++)
        ans = max(ans * 1LL, dp[v][elim] - elim * k);
      sol[tc] = ans;
    }
  }
  for (int i = 1; i <= q; i++) cout << sol[i] << '\n';
  return 0;
}
