#include <bits/stdc++.h>
using namespace std;
namespace Ruri {
const int infi = 1e9;
const long long infl = 4e18;
}  // namespace Ruri
using namespace Ruri;
namespace Gadgets {
inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
inline long long pos_mod(long long a, long long b) { return (a % b + b) % b; }
}  // namespace Gadgets
namespace Read {
inline int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long rl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline double rd() {
  double x;
  scanf("%lf", &x);
  return x;
}
}  // namespace Read
namespace DeBug {}
using namespace Gadgets;
using namespace Read;
using namespace DeBug;
const int MAX_N = 2e5 + 5;
int n;
vector<int> g[MAX_N];
vector<pair<int, int> > query[MAX_N];
int son[MAX_N];
int ans[MAX_N];
vector<int> dp[MAX_N];
void Dfs(int u, int fa) {
  dp[u].resize(son[u]);
  for (auto v : g[u])
    if (v != fa) {
      Dfs(v, u);
      if ((int)dp[u].size() < (int)dp[v].size())
        dp[u].resize((int)dp[v].size());
      for (int i = 0, bbb = (int)dp[v].size() - 1; i <= bbb; ++i)
        dp[u][i] += dp[v][i] - 1;
    }
  for (auto x : query[u])
    if (x.second < (int)dp[u].size()) ans[x.first] += dp[u][x.second];
  for (int i = 0, bbb = (int)dp[u].size() - 1; i <= bbb; ++i)
    dp[u][i] += son[u] - i;
  while (!dp[u].empty() && dp[u].back() <= 0) dp[u].pop_back();
}
int main() {
  n = ri();
  for (int i = 1, bbb = n - 1; i <= bbb; ++i) {
    int u = ri(), v = ri();
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 1, bbb = n; i <= bbb; ++i) son[i] = (int)g[i].size() - (i != 1);
  int q = ri();
  for (int i = 1, bbb = q; i <= bbb; ++i) {
    int v = ri(), k = ri();
    query[v].push_back(pair<int, int>(i, k));
    ans[i] = son[v];
  }
  Dfs(1, 0);
  for (int i = 1, bbb = q; i <= bbb; ++i) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Please create a solution in cpp 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>
using namespace std;
namespace Ruri {
const int infi = 1e9;
const long long infl = 4e18;
}  // namespace Ruri
using namespace Ruri;
namespace Gadgets {
inline long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
inline long long pos_mod(long long a, long long b) { return (a % b + b) % b; }
}  // namespace Gadgets
namespace Read {
inline int ri() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long rl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
inline double rd() {
  double x;
  scanf("%lf", &x);
  return x;
}
}  // namespace Read
namespace DeBug {}
using namespace Gadgets;
using namespace Read;
using namespace DeBug;
const int MAX_N = 2e5 + 5;
int n;
vector<int> g[MAX_N];
vector<pair<int, int> > query[MAX_N];
int son[MAX_N];
int ans[MAX_N];
vector<int> dp[MAX_N];
void Dfs(int u, int fa) {
  dp[u].resize(son[u]);
  for (auto v : g[u])
    if (v != fa) {
      Dfs(v, u);
      if ((int)dp[u].size() < (int)dp[v].size())
        dp[u].resize((int)dp[v].size());
      for (int i = 0, bbb = (int)dp[v].size() - 1; i <= bbb; ++i)
        dp[u][i] += dp[v][i] - 1;
    }
  for (auto x : query[u])
    if (x.second < (int)dp[u].size()) ans[x.first] += dp[u][x.second];
  for (int i = 0, bbb = (int)dp[u].size() - 1; i <= bbb; ++i)
    dp[u][i] += son[u] - i;
  while (!dp[u].empty() && dp[u].back() <= 0) dp[u].pop_back();
}
int main() {
  n = ri();
  for (int i = 1, bbb = n - 1; i <= bbb; ++i) {
    int u = ri(), v = ri();
    g[u].push_back(v), g[v].push_back(u);
  }
  for (int i = 1, bbb = n; i <= bbb; ++i) son[i] = (int)g[i].size() - (i != 1);
  int q = ri();
  for (int i = 1, bbb = q; i <= bbb; ++i) {
    int v = ri(), k = ri();
    query[v].push_back(pair<int, int>(i, k));
    ans[i] = son[v];
  }
  Dfs(1, 0);
  for (int i = 1, bbb = q; i <= bbb; ++i) printf("%d\n", ans[i]);
  return 0;
}

```