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;
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename C, typename T = decay<decltype(*begin(declval<C>()))>,
          typename enable_if<!is_same<C, string>::value>::type * = nullptr>
ostream &operator<<(ostream &os, const C &c) {
  bool f = true;
  os << "[";
  for (const auto &x : c) {
    if (!f) os << ", ";
    f = false;
    os << x;
  }
  return os << "]";
}
template <typename T>
void debug(string s, T x) {
  cerr << "\033[1;35m" << s << "\033[0;32m = \033[33m" << x << "\033[0m\n";
}
template <typename T, typename... Args>
void debug(string s, T x, Args... args) {
  for (int i = 0, b = 0; i < (int)s.size(); i++)
    if (s[i] == '(' || s[i] == '{')
      b++;
    else if (s[i] == ')' || s[i] == '}')
      b--;
    else if (s[i] == ',' && b == 0) {
      cerr << "\033[1;35m" << s.substr(0, i) << "\033[0;32m = \033[33m" << x
           << "\033[31m | ";
      debug(s.substr(s.find_first_not_of(' ', i + 1)), args...);
      break;
    }
}
const int MAX = 2e5 + 5;
const int SZ = 450;
int dp[MAX][SZ], ndp[SZ], sz[MAX], ret[MAX];
vector<int> adj[MAX];
vector<array<int, 3>> queries[2];
void dfs1(int u, int p) {
  dp[u][0] = 0;
  sz[u] = 1;
  for (int v : adj[u])
    if (v != p) {
      dfs1(v, u);
      memcpy(ndp, dp[u], sizeof(int) * sz[u]);
      for (int i = sz[u]; i < min(sz[u] + sz[v], SZ); i++) ndp[i] = 0;
      ndp[0]++;
      for (int i = 0; i < sz[u]; i++)
        for (int j = 0; j < min(sz[v], SZ - sz[u]); j++)
          ndp[i + j + 1] = max(ndp[i + j + 1], dp[u][i] + dp[v][j]);
      sz[u] = min(sz[u] + sz[v], SZ);
      memcpy(dp[u], ndp, sizeof(int) * sz[u]);
    }
}
void dfs2(int u, int p) {
  for (int v : adj[u])
    if (v != p) {
      dfs2(v, u);
      for (int k = 0; k < SZ; k++) dp[u][k] += max(dp[v][k] - k, 1);
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v, k;
    cin >> v >> k;
    v--;
    queries[k < SZ].push_back({v, k, i});
  }
  dfs1(0, -1);
  for (auto [v, k, i] : queries[0])
    for (int j = 0; j < sz[v]; j++)
      ret[i] = (int)max((long long)ret[i], dp[v][j] - (long long)j * k);
  memset(dp, 0, sizeof(dp));
  dfs2(0, -1);
  for (auto [v, k, i] : queries[1]) ret[i] = dp[v][k];
  for (int i = 0; i < q; i++) cout << ret[i] << "\n";
  return 0;
}
