#include <bits/stdc++.h>
using namespace std;
const int sqn = 58, inf = 1e9;
int n, eu[200069], pr[200069], dp[200069], dp2[200069][69], tmp[69], tmp2[69],
    sq[200069];
vector<int> al[200069], ql[200069];
bitset<200069> vtd;
pair<int, int> qq[200069];
void dfs(int x) {
  int i, sz = al[x].size(), l;
  vtd[x] = 1;
  n++;
  eu[n] = x;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    if (!vtd[l]) {
      pr[l] = x;
      dfs(l);
    }
  }
}
int main() {
  int t, rr, i, j, r, k, l, sz, pz;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &k, &l);
    al[k].push_back(l);
    al[l].push_back(k);
  }
  n = 0;
  dfs(1);
  scanf("%d", &t);
  for (rr = 1; rr <= t; rr++) {
    scanf("%d%d", &k, &l);
    qq[rr] = {k, l};
    ql[min(l, n / sqn + 1)].push_back(rr);
  }
  for (i = 0; i <= n / sqn; i++) {
    for (j = 1; j <= n; j++) {
      dp[j] = 0;
    }
    for (j = n; j; j--) {
      k = eu[j];
      dp[pr[k]] += max(dp[k] - i, 1);
    }
    sz = ql[i].size();
    for (j = 0; j < sz; j++) {
      pz = ql[i][j];
      k = qq[pz].first;
      sq[pz] = dp[k];
    }
  }
  for (i = n; i; i--) {
    k = eu[i];
    for (j = 0; j <= sqn; j++) {
      if (!j) {
        tmp[j] = 1;
      } else {
        tmp[j] = dp2[k][j - 1];
      }
      tmp2[j] = -inf;
    }
    for (j = 0; j <= sqn; j++) {
      for (r = 0; r <= sqn - j; r++) {
        tmp2[j + r] = max(tmp2[j + r], dp2[pr[k]][j] + tmp[r]);
      }
    }
    for (j = 0; j <= sqn; j++) {
      dp2[pr[k]][j] = tmp2[j];
    }
  }
  sz = ql[n / sqn + 1].size();
  for (i = 0; i < sz; i++) {
    pz = ql[n / sqn + 1][i];
    k = qq[pz].first;
    l = qq[pz].second;
    for (j = 0; j <= sqn; j++) {
      sq[pz] = max(sq[pz], dp2[k][j] - j * l);
    }
  }
  for (rr = 1; rr <= t; rr++) {
    printf("%d\n", sq[rr]);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int sqn = 58, inf = 1e9;
int n, eu[200069], pr[200069], dp[200069], dp2[200069][69], tmp[69], tmp2[69],
    sq[200069];
vector<int> al[200069], ql[200069];
bitset<200069> vtd;
pair<int, int> qq[200069];
void dfs(int x) {
  int i, sz = al[x].size(), l;
  vtd[x] = 1;
  n++;
  eu[n] = x;
  for (i = 0; i < sz; i++) {
    l = al[x][i];
    if (!vtd[l]) {
      pr[l] = x;
      dfs(l);
    }
  }
}
int main() {
  int t, rr, i, j, r, k, l, sz, pz;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &k, &l);
    al[k].push_back(l);
    al[l].push_back(k);
  }
  n = 0;
  dfs(1);
  scanf("%d", &t);
  for (rr = 1; rr <= t; rr++) {
    scanf("%d%d", &k, &l);
    qq[rr] = {k, l};
    ql[min(l, n / sqn + 1)].push_back(rr);
  }
  for (i = 0; i <= n / sqn; i++) {
    for (j = 1; j <= n; j++) {
      dp[j] = 0;
    }
    for (j = n; j; j--) {
      k = eu[j];
      dp[pr[k]] += max(dp[k] - i, 1);
    }
    sz = ql[i].size();
    for (j = 0; j < sz; j++) {
      pz = ql[i][j];
      k = qq[pz].first;
      sq[pz] = dp[k];
    }
  }
  for (i = n; i; i--) {
    k = eu[i];
    for (j = 0; j <= sqn; j++) {
      if (!j) {
        tmp[j] = 1;
      } else {
        tmp[j] = dp2[k][j - 1];
      }
      tmp2[j] = -inf;
    }
    for (j = 0; j <= sqn; j++) {
      for (r = 0; r <= sqn - j; r++) {
        tmp2[j + r] = max(tmp2[j + r], dp2[pr[k]][j] + tmp[r]);
      }
    }
    for (j = 0; j <= sqn; j++) {
      dp2[pr[k]][j] = tmp2[j];
    }
  }
  sz = ql[n / sqn + 1].size();
  for (i = 0; i < sz; i++) {
    pz = ql[n / sqn + 1][i];
    k = qq[pz].first;
    l = qq[pz].second;
    for (j = 0; j <= sqn; j++) {
      sq[pz] = max(sq[pz], dp2[k][j] - j * l);
    }
  }
  for (rr = 1; rr <= t; rr++) {
    printf("%d\n", sq[rr]);
  }
}

```