#include <bits/stdc++.h>
using namespace std;
long long cc = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<vector<int>> dp(n);
  vector<int> sz(n);
  function<void(int, int)> dfs = [&](int u, int fa) {
    sz[u] = 1;
    int son = 0;
    vector<int> res;
    for (int v : g[u]) {
      if (v == fa) continue;
      dfs(v, u);
      sz[u] += sz[v];
      son += 1;
      for (int k = 0;; ++k) {
        if (k == dp[u].size()) dp[u].push_back(0);
        int p = k < dp[v].size() ? dp[v][k] - k : dp[v].back() - k;
        if (p <= 0) {
          res.push_back(k);
          break;
        }
        dp[u][k] += p;
      }
    }
    if (dp[u].empty()) dp[u].push_back(0);
    int m = dp[u].size();
    cc += m;
    vector<int> c(m + 1);
    for (int i : res) c[i] += 1;
    for (int i = 1; i < m; ++i) c[i] += c[i - 1];
    for (int i = 0; i < m; ++i) dp[u][i] += c[i];
  };
  assert(cc < n);
  dfs(0, -1);
  int q;
  cin >> q;
  while (q--) {
    int u, k;
    cin >> u >> k;
    --u;
    if (k < dp[u].size())
      cout << dp[u][k] << '\n';
    else
      cout << dp[u].back() << '\n';
  }
}

### Prompt

In cpp, your task is to solve the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long cc = 0;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> g(n);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  vector<vector<int>> dp(n);
  vector<int> sz(n);
  function<void(int, int)> dfs = [&](int u, int fa) {
    sz[u] = 1;
    int son = 0;
    vector<int> res;
    for (int v : g[u]) {
      if (v == fa) continue;
      dfs(v, u);
      sz[u] += sz[v];
      son += 1;
      for (int k = 0;; ++k) {
        if (k == dp[u].size()) dp[u].push_back(0);
        int p = k < dp[v].size() ? dp[v][k] - k : dp[v].back() - k;
        if (p <= 0) {
          res.push_back(k);
          break;
        }
        dp[u][k] += p;
      }
    }
    if (dp[u].empty()) dp[u].push_back(0);
    int m = dp[u].size();
    cc += m;
    vector<int> c(m + 1);
    for (int i : res) c[i] += 1;
    for (int i = 1; i < m; ++i) c[i] += c[i - 1];
    for (int i = 0; i < m; ++i) dp[u][i] += c[i];
  };
  assert(cc < n);
  dfs(0, -1);
  int q;
  cin >> q;
  while (q--) {
    int u, k;
    cin >> u >> k;
    --u;
    if (k < dp[u].size())
      cout << dp[u][k] << '\n';
    else
      cout << dp[u].back() << '\n';
  }
}

```