#include <bits/stdc++.h>
using namespace std;
mt19937_64 rnd(time(NULL));
const int N = 1000010;
const long long mod = 998244353;
vector<int> G[N];
vector<int> g[N];
int p[N];
int tin[N], tout[N];
int timer = 0;
void dfs0(int v, int pr) {
  p[v] = pr;
  tin[v] = ++timer;
  for (int to : G[v]) {
    if (to != pr) {
      g[v].push_back(to);
      dfs0(to, v);
    }
  }
  tout[v] = timer;
}
int n, q;
int ans[N];
bool good[N];
int dsu[N];
int get(int x) {
  if (!good[p[dsu[x]]]) {
    return dsu[x];
  }
  return dsu[x] = get(p[dsu[x]]);
}
int val[N];
int sum[N], cnt[N];
struct BIT {
  int T[N];
  void upd(int pos, int x) {
    for (; pos <= n; pos |= pos + 1) T[pos] += x;
  }
  int get(int pos) {
    int sum = 0;
    for (; pos >= 0; pos &= pos + 1, pos--) sum += T[pos];
    return sum;
  }
};
BIT T[2];
void add(int l, int r, int x, int i) {
  T[i].upd(l, x);
  T[i].upd(r + 1, -x);
}
void solve() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  cin >> q;
  dfs0(1, 0);
  vector<pair<pair<int, int>, int> > Q;
  for (int i = 1; i <= q; i++) {
    int v, k;
    cin >> v >> k;
    Q.push_back({{k, v}, i});
  }
  sort(Q.begin(), Q.end());
  reverse(Q.begin(), Q.end());
  for (int i = 1; i <= n; i++)
    good[i] = false, dsu[i] = i, sum[i] = g[i].size() - 1, cnt[i] = 1;
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) st.insert({sum[i] / cnt[i], i});
  for (auto cur : Q) {
    int k = cur.first.first;
    int v = cur.first.second;
    while (!st.empty()) {
      int v = (*st.rbegin()).second;
      if (sum[v] / cnt[v] < k) break;
      st.erase(--st.end());
      good[v] = true;
      int pr = p[get(v)];
      T[0].upd(tin[p[pr]], -sum[v]);
      T[0].upd(tin[p[v]], sum[v]);
      T[1].upd(tin[p[pr]], -cnt[v]);
      T[1].upd(tin[p[v]], cnt[v]);
      if (pr) {
        st.erase({sum[pr] / cnt[pr], pr});
        sum[pr] += sum[v];
        cnt[pr] += cnt[v];
        st.insert({sum[pr] / cnt[pr], pr});
      }
    }
    int res = g[v].size() + T[0].get(tout[v]) - T[0].get(tin[v] - 1) -
              (T[1].get(tout[v]) - T[1].get(tin[v] - 1)) * k;
    ans[cur.second] = res;
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  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;
mt19937_64 rnd(time(NULL));
const int N = 1000010;
const long long mod = 998244353;
vector<int> G[N];
vector<int> g[N];
int p[N];
int tin[N], tout[N];
int timer = 0;
void dfs0(int v, int pr) {
  p[v] = pr;
  tin[v] = ++timer;
  for (int to : G[v]) {
    if (to != pr) {
      g[v].push_back(to);
      dfs0(to, v);
    }
  }
  tout[v] = timer;
}
int n, q;
int ans[N];
bool good[N];
int dsu[N];
int get(int x) {
  if (!good[p[dsu[x]]]) {
    return dsu[x];
  }
  return dsu[x] = get(p[dsu[x]]);
}
int val[N];
int sum[N], cnt[N];
struct BIT {
  int T[N];
  void upd(int pos, int x) {
    for (; pos <= n; pos |= pos + 1) T[pos] += x;
  }
  int get(int pos) {
    int sum = 0;
    for (; pos >= 0; pos &= pos + 1, pos--) sum += T[pos];
    return sum;
  }
};
BIT T[2];
void add(int l, int r, int x, int i) {
  T[i].upd(l, x);
  T[i].upd(r + 1, -x);
}
void solve() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  cin >> q;
  dfs0(1, 0);
  vector<pair<pair<int, int>, int> > Q;
  for (int i = 1; i <= q; i++) {
    int v, k;
    cin >> v >> k;
    Q.push_back({{k, v}, i});
  }
  sort(Q.begin(), Q.end());
  reverse(Q.begin(), Q.end());
  for (int i = 1; i <= n; i++)
    good[i] = false, dsu[i] = i, sum[i] = g[i].size() - 1, cnt[i] = 1;
  set<pair<int, int> > st;
  for (int i = 1; i <= n; i++) st.insert({sum[i] / cnt[i], i});
  for (auto cur : Q) {
    int k = cur.first.first;
    int v = cur.first.second;
    while (!st.empty()) {
      int v = (*st.rbegin()).second;
      if (sum[v] / cnt[v] < k) break;
      st.erase(--st.end());
      good[v] = true;
      int pr = p[get(v)];
      T[0].upd(tin[p[pr]], -sum[v]);
      T[0].upd(tin[p[v]], sum[v]);
      T[1].upd(tin[p[pr]], -cnt[v]);
      T[1].upd(tin[p[v]], cnt[v]);
      if (pr) {
        st.erase({sum[pr] / cnt[pr], pr});
        sum[pr] += sum[v];
        cnt[pr] += cnt[v];
        st.insert({sum[pr] / cnt[pr], pr});
      }
    }
    int res = g[v].size() + T[0].get(tout[v]) - T[0].get(tin[v] - 1) -
              (T[1].get(tout[v]) - T[1].get(tin[v] - 1)) * k;
    ans[cur.second] = res;
  }
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}

```