#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimize(0)
#pragma GCC optimization("unroll-loops")
using namespace std;
const int MAXN = 1e3 + 1, inf = 1e9 + 7;
int M = 1e9 + 7, P = 1e6 + 2;
int f[MAXN + 1], inv[MAXN + 1], inc[MAXN + 1];
int bin_pow(int a, int b, int M) {
  if (b <= 0) return 1;
  if (b % 2 == 0) {
    int x = bin_pow(a, b / 2, M);
    return x * x % M;
  } else {
    return bin_pow(a, b - 1, M) * a % M;
  }
}
int C(int n, int k) {
  if (n < 0 || k > n) return 0;
  return (f[n] * inc[k] % M * inc[n - k] % M + M) % M;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) {
  if (a < 0) return lcm(abs(a), b);
  if (b < 0) return lcm(a, abs(b));
  if (a == 0 && b == 0) return 0;
  return a / gcd(a, b) * b;
}
vector<vector<int>> g, dp;
int get(int u, int k) {
  k = min(k, (int)dp[u].size() - 1);
  return dp[u][k];
}
void dfs(int u, int p) {
  int cnt = 0;
  vector<int> to;
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u);
      cnt++;
      to.push_back(v);
    }
  }
  int s = 0;
  for (int i = 0; i < (int)g.size(); ++i) {
    int tmp = 0;
    for (int v : to) {
      tmp += max((int)1, get(v, i) - i);
    }
    dp[u].push_back(tmp + s);
    if (i == 250) {
      vector<int> to_2;
      for (int v : to) {
        if (get(v, i) - i < 2)
          ++s;
        else
          to_2.push_back(v);
      }
      swap(to, to_2);
    }
    if (tmp + s == cnt) break;
  }
}
void solve() {
  int n, m;
  cin >> n;
  g.resize(n);
  dp.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  int q;
  cin >> q;
  while (q--) {
    int u, k;
    cin >> u >> k;
    --u;
    cout << get(u, k) << '\n';
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  f[0] = inv[0] = f[1] = inv[1] = 1;
  for (int i = 2; i <= MAXN; ++i) {
    f[i] = f[i - 1] * i % M;
    inv[i] = ((-M / i) * inv[M % i] + M) % M;
  }
  inc[0] = 1;
  for (int i = 1; i <= MAXN; ++i) {
    inc[i] = inc[i - 1] * inv[i] % M;
  }
  int _ = 1;
  while (_--) solve();
}

### Prompt

Your challenge is to write a CPP solution to 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>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimize(0)
#pragma GCC optimization("unroll-loops")
using namespace std;
const int MAXN = 1e3 + 1, inf = 1e9 + 7;
int M = 1e9 + 7, P = 1e6 + 2;
int f[MAXN + 1], inv[MAXN + 1], inc[MAXN + 1];
int bin_pow(int a, int b, int M) {
  if (b <= 0) return 1;
  if (b % 2 == 0) {
    int x = bin_pow(a, b / 2, M);
    return x * x % M;
  } else {
    return bin_pow(a, b - 1, M) * a % M;
  }
}
int C(int n, int k) {
  if (n < 0 || k > n) return 0;
  return (f[n] * inc[k] % M * inc[n - k] % M + M) % M;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int lcm(int a, int b) {
  if (a < 0) return lcm(abs(a), b);
  if (b < 0) return lcm(a, abs(b));
  if (a == 0 && b == 0) return 0;
  return a / gcd(a, b) * b;
}
vector<vector<int>> g, dp;
int get(int u, int k) {
  k = min(k, (int)dp[u].size() - 1);
  return dp[u][k];
}
void dfs(int u, int p) {
  int cnt = 0;
  vector<int> to;
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u);
      cnt++;
      to.push_back(v);
    }
  }
  int s = 0;
  for (int i = 0; i < (int)g.size(); ++i) {
    int tmp = 0;
    for (int v : to) {
      tmp += max((int)1, get(v, i) - i);
    }
    dp[u].push_back(tmp + s);
    if (i == 250) {
      vector<int> to_2;
      for (int v : to) {
        if (get(v, i) - i < 2)
          ++s;
        else
          to_2.push_back(v);
      }
      swap(to, to_2);
    }
    if (tmp + s == cnt) break;
  }
}
void solve() {
  int n, m;
  cin >> n;
  g.resize(n);
  dp.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0, -1);
  int q;
  cin >> q;
  while (q--) {
    int u, k;
    cin >> u >> k;
    --u;
    cout << get(u, k) << '\n';
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  f[0] = inv[0] = f[1] = inv[1] = 1;
  for (int i = 2; i <= MAXN; ++i) {
    f[i] = f[i - 1] * i % M;
    inv[i] = ((-M / i) * inv[M % i] + M) % M;
  }
  inc[0] = 1;
  for (int i = 1; i <= MAXN; ++i) {
    inc[i] = inc[i - 1] * inv[i] % M;
  }
  int _ = 1;
  while (_--) solve();
}

```