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("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
using namespace std;
const int MAX = 2e5 + 2;
const int SQRT = 448;
int n;
vector<vector<int> > adj(MAX);
int dp[MAX][SQRT + 2];
int tmpdp[SQRT + 2];
void dfs(int node, int par) {
  for (auto &child : adj[node]) {
    if (child == par) continue;
    dfs(child, node);
    for (int j = 0; j <= SQRT; ++j) dp[node][j] += max(1, dp[child][j] - j);
  }
}
void dfs2(int node, int par) {
  for (auto &child : adj[node]) {
    if (child == par) continue;
    dfs2(child, node);
  }
  memset(tmpdp, 0, sizeof(tmpdp));
  for (auto &child : adj[node]) {
    int Max = 0;
    for (int j = 0; j <= SQRT; ++j) {
      if (!dp[node][j] && j) break;
      for (int j2 = 0; j + j2 + 1 <= SQRT; ++j2) {
        if (!dp[child][j2]) break;
        Max = max(Max, j + j2 + 1);
        tmpdp[j + j2 + 1] =
            max(tmpdp[j + j2 + 1],
                max(dp[node][j + j2 + 1] + 1, dp[node][j] + dp[child][j2]));
      }
    }
    for (int i = 0; i <= Max; ++i) {
      dp[node][i] = tmpdp[i];
      tmpdp[i]++;
    }
  }
}
int qu[MAX], qk[MAX], ans[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, -1);
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    int node, k;
    cin >> node >> k;
    qu[i] = node, qk[i] = k;
    if (k <= SQRT) ans[i] = dp[node][k];
  }
  memset(dp, 0, sizeof(dp));
  dfs2(1, -1);
  for (int i = 1; i <= q; ++i) {
    int node = qu[i], k = qk[i];
    if (k > SQRT) {
      for (int j = 0; j <= SQRT; ++j) ans[i] = max(ans[i], dp[node][j] - k * j);
    }
    cout << ans[i] << "\n";
  }
  return 0;
}
