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;
int n, m, q, c[200015], ans[200015], dep[200015], par[200015];
vector<int> e[200015];
vector<pair<int, int> > que[200015];
int fa[200015], siz[200015], rt[200015], tc[200015], tk[200015];
void init() {
  for (int i = (1); i <= (n); i++) fa[i] = rt[i] = i, siz[i] = 1, tc[i] = c[i];
}
int find(int x) { return x == fa[x] ? x : find(fa[x]); }
void uni(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return;
  if (siz[u] < siz[v]) swap(u, v);
  fa[v] = u;
  tc[v] -= tc[u];
  tk[v] -= tk[u];
  siz[u] += siz[v];
  rt[u] = dep[rt[v]] < dep[rt[u]] ? rt[v] : rt[u];
}
int qc(int x) {
  int res = 0;
  while (x != fa[x]) {
    res += tc[x];
    x = fa[x];
  }
  res += tc[x];
  return res;
}
int qk(int x) {
  int res = 0;
  while (x != fa[x]) {
    res += tk[x];
    x = fa[x];
  }
  res += tk[x];
  return res;
}
void addc(int x, int v) { tc[x] += v; }
void addk(int x, int v) { tk[x] += v; }
void dfs(int u, int pa) {
  dep[u] = dep[pa] + 1;
  par[u] = pa;
  for (auto v : e[u])
    if (v != pa) {
      c[u]++;
      dfs(v, u);
    }
}
set<pair<int, int> > S;
int main() {
  scanf("%d", &n);
  for (int i = (2); i <= (n); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  scanf("%d", &q);
  for (int i = (1); i <= (q); i++) {
    int v, k;
    scanf("%d%d", &v, &k);
    m = max(m, k);
    que[k].push_back(make_pair(v, i));
  }
  dfs(1, 0);
  init();
  for (int i = (1); i <= (n); i++) S.insert(make_pair(-c[i], i));
  for (int i = (m); i >= (0); i--) {
    while (((int)S.size()) > 0) {
      if (-S.begin()->first <= i) break;
      int u = S.begin()->second;
      S.erase(S.begin());
      if (!par[u]) continue;
      int v = find(par[u]);
      assert(u == rt[find(u)]);
      S.erase(make_pair(-qc(rt[v]), rt[v]));
      addc(v, qc(u) - 1);
      addk(v, 1);
      uni(u, v);
      if (par[rt[v]]) S.insert(make_pair(-qc(rt[v]), rt[v]));
    }
    for (auto x : que[i]) {
      ans[x.second] = qc(x.first) - qk(x.first) * i;
    }
  }
  for (int i = (1); i <= (q); i++) printf("%d\n", ans[i]);
  return 0;
}
