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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
mt19937 rnf(2106);
const int N = 200005, S = 440;
int n;
vector<int> g[N];
int d[N];
int u[N];
int t[N];
bool sot(int x, int y) { return t[x] > t[y]; }
void dfs00(int x, int p) {
  if (x == p)
    d[x] = 0;
  else
    d[x] = d[p] + 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) continue;
    dfs00(h, x);
  }
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (h == p) {
      swap(g[x][i], g[x].back());
      g[x].pop_back();
      break;
    }
  }
  u[x] = x;
  int q = 0;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (t[h] >= S) {
      ++q;
    }
  }
  if (q == 1) {
    for (int i = 0; i < g[x].size(); ++i) {
      int h = g[x][i];
      if (t[h] >= S) {
        u[x] = u[h];
        break;
      }
    }
  }
  if (g[x].empty())
    t[x] = 1;
  else {
    for (int i = 0; i < g[x].size(); ++i) {
      int h = g[x][i];
      t[x] += t[h];
    }
  }
}
int dp0[N][S];
void dfs0(int x) {
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    dfs0(h);
    for (int k = 0; k < S; ++k) dp0[x][k] += max(1, dp0[h][k] - k);
  }
}
int dp[N];
void dfs(int x, int k) {
  dp[x] = 0;
  for (int i = 0; i < g[x].size(); ++i) {
    int h = g[x][i];
    if (t[h] <= k) {
      dp[x] += (((int)(g[x]).size()) - i);
      break;
    }
    dfs(u[h], k);
    dp[x] += max(1LL, dp[u[h]] + (d[x] - d[u[h]]) * 1LL * k);
  }
}
void solv() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs00(1, 1);
  dfs0(1);
  for (int x = 1; x <= n; ++x) sort((g[x]).begin(), (g[x]).end(), sot);
  int qq;
  cin >> qq;
  while (qq--) {
    int x, k;
    cin >> x >> k;
    if (k < S) {
      cout << dp0[x][k] << "\n";
      continue;
    }
    dfs(x, k);
    cout << dp[x] << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  int tt = 1;
  while (tt--) solv();
  return 0;
}
