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 MOD = (int)1e9 + 7;
const int modulo = 998244353;
;
const int INF = (int)1e9 + 47;
const int nax = 2 * (int)1e5 + 10;
const int lim = 2 * (int)1e3 + 10;
const double EPS = 1e-10;
const int sqrt_const = 451;
int v[nax], k[nax];
int dp[nax][sqrt_const];
int sz[nax];
int cur[nax];
vector<int> g[nax];
int ans[nax];
void small(int v, int p = -1) {
  for (auto to : g[v]) {
    if (to == p) continue;
    small(to, v);
    for (int magik = 0; magik < sqrt_const; magik++) {
      dp[v][magik] += max(dp[to][magik] - magik, 1);
    }
  }
}
void big(int v, int p = -1) {
  sz[v] = 1;
  for (int i = 0; i < sqrt_const; i++) dp[v][i] = 0;
  for (auto to : g[v]) {
    if (to == p) continue;
    big(to, v);
    for (int i = 0; i < sqrt_const; i++) cur[i] = 0;
    for (int i = 0; i < min(sz[v] + 1, sqrt_const); i++) {
      cur[i] = max(cur[i], dp[v][i] + 1);
      for (int j = 0; j <= sz[to]; j++) {
        if (i + j + 1 >= sqrt_const) break;
        cur[i + j + 1] = max(cur[i + j + 1], dp[v][i] + dp[v][j]);
      }
    }
    sz[v] += sz[to];
    for (int i = 0; i < min(sz[v] + 1, sqrt_const); i++) dp[v][i] = cur[i];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int q;
  cin >> q;
  for (int qq = 0; qq < q; qq++) {
    cin >> v[qq] >> k[qq];
    --v[qq];
  }
  small(0, -1);
  for (int qq = 0; qq < q; qq++) {
    if (k[qq] < sqrt_const) ans[qq] = dp[v[qq]][k[qq]];
  }
  big(0, -1);
  for (int qq = 0; qq < q; qq++) {
    if (k[qq] >= sqrt_const) {
      for (int j = 0; j < sqrt_const; j++)
        ans[qq] = max(ans[qq], dp[v[qq]][j] - j * k[qq]);
    }
  }
  for (int qq = 0; qq < q; qq++) {
    cout << ans[qq] << '\n';
  }
}
