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;
const int maxN = 2e5 + 10;
vector<int> g[maxN];
const int BUBEN = 350;
int q;
long long ans[maxN];
vector<int> st[BUBEN + 2];
int k[maxN], vv[maxN];
long long dp[maxN];
vector<int> ord;
vector<int> edge[maxN];
int par[maxN];
int tin[maxN];
int timer = 0;
int tout[maxN];
int h[maxN];
void dfs(int v, int p) {
  tin[v] = timer++;
  for (int to : g[v]) {
    if (to == p) continue;
    par[to] = v;
    edge[v].emplace_back(to);
    h[to] = h[v] + 1;
    dfs(to, v);
  }
  ord.emplace_back(v);
  tout[v] = timer++;
}
int bigPar[maxN];
vector<pair<int, int>> big_edge[maxN];
bool byTin(int a, int b) { return tin[a] < tin[b]; }
int lca(int x, int y) {
  while (h[x] > h[y]) {
    x = par[x];
  }
  while (h[x] < h[y]) {
    y = par[y];
  }
  while (x != y) {
    x = par[x];
    y = par[y];
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  dfs(1, -1);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> vv[i] >> k[i];
    if (k[i] <= BUBEN) {
      st[k[i]].emplace_back(i);
    }
  }
  for (int i = 0; i <= BUBEN; i++) {
    if (!st[i].empty()) {
      for (int vert : ord) {
        dp[vert] = 0;
        for (int x : edge[vert]) {
          dp[vert] += max(1LL, dp[x] - i);
        }
      }
      for (int id : st[i]) {
        ans[id] = dp[vv[id]];
      }
    }
  }
  vector<int> inter;
  for (int z : ord) {
    if ((int)edge[z].size() > BUBEN || z == 1) {
      inter.emplace_back(z);
    }
  }
  sort(inter.begin(), inter.end(), byTin);
  int m = inter.size();
  for (int i = 0; i + 1 < m; i++) {
    inter.emplace_back(lca(inter[i], inter[i + 1]));
  }
  sort(inter.begin(), inter.end(), byTin);
  inter.erase(unique(inter.begin(), inter.end()), inter.end());
  vector<int> st;
  st.push_back(inter[0]);
  for (int i = 1; i < inter.size(); i++) {
    int x = inter[i];
    while (tout[st.back()] < tin[x]) st.pop_back();
    int y = st.back();
    big_edge[y].emplace_back(make_pair(x, -h[y] + h[x]));
    bigPar[x] = y;
    st.push_back(x);
  }
  reverse(inter.begin(), inter.end());
  for (int i = 1; i <= q; i++) {
    if (k[i] > BUBEN) {
      int who = vv[i];
      vector<int> without_par;
      bool hs = false;
      for (int u : inter) {
        if (u == who) hs = true;
        if (tin[who] <= tin[u] && tout[u] <= tout[who]) {
          if (u == 1 ||
              !(tin[who] <= tin[bigPar[u]] && tout[bigPar[u]] <= tout[who])) {
            without_par.emplace_back(u);
          }
          dp[u] = 0;
          for (auto& it : big_edge[u]) {
            dp[u] += max(1LL, dp[it.first] - 1LL * it.second * k[i]);
          }
        }
      }
      if (hs) {
        cout << dp[who] << '\n';
      } else {
        dp[who] = 0;
        for (int z : without_par) {
          dp[who] += max(1LL, dp[z] - 1LL * k[i] * (h[z] - h[i]));
        }
        dp[who] += edge[who].size() - without_par.size();
        cout << dp[who] << '\n';
      }
    } else {
      cout << ans[i] << '\n';
    }
  }
  return 0;
}
