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 N = 2e5 + 99, sq = 500;
int n, q, sz[N], ans[N], a[N][2], dp[N][sq];
vector<int> g[N];
void dfs1(int x, int par) {
  for (int i = 0; i < g[x].size(); i++) {
    if (g[x][i] != par) {
      dfs1(g[x][i], x);
      for (int j = 0; j < sq; j++) {
        dp[x][j] += max(1, dp[g[x][i]][j] - j);
      }
    }
  }
}
void dfs2(int x, int par) {
  sz[x] = 1;
  dp[x][0] = -N;
  for (int i = 1; i < sq; i++) dp[x][i] = 0;
  for (int i = 0; i < g[x].size(); i++) {
    if (g[x][i] != par) {
      dfs2(g[x][i], x);
      for (int j = sz[x] / sq + 1; j >= 0; j--) {
        for (int k = sz[g[x][i]] / sq + 1; k >= 0; k--) {
          dp[x][j + k] = max(dp[x][j + k], dp[x][j] + dp[g[x][i]][k]);
        }
      }
      sz[x] += sz[g[x][i]];
    }
  }
  dp[x][0] = 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    u = i, v = i + 1;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs1(1, 0);
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> a[i][0] >> a[i][1];
    if (a[i][1] < sq) {
      ans[i] = dp[a[i][0]][a[i][1]];
    }
  }
  dfs2(1, 0);
  for (int i = 0; i < q; i++) {
    if (a[i][1] >= sq) {
      int res = 0, u = a[i][0], x = a[i][1];
      for (int j = 1; j < sq; j++) {
        res = max(res, dp[u][j] - (j - 1) * x);
      }
      ans[i] = res;
    }
  }
  for (int i = 0; i < q; i++) {
    cout << ans[i] << endl;
  }
}
