#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
using namespace std;
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 1e9 + 100;
const int mod = 1000000007;
const long double eps = 1e-11, pi = acos(-1);
const long long maxN = 200010, maxT = 1010, A = 179, K = 200;
mt19937 mt_rand(time(0));
long long bp(long long et, long long b) {
  b %= mod - 1;
  et %= mod;
  long long res = 1;
  for (int i = 30; i >= 0; --i) {
    res = (res * res) % mod;
    if ((b & (1 << i)) != 0) res = (res * et) % mod;
  }
  return res;
}
void panic() {
  cout << "impossible\n";
  exit(0);
}
int f[K][maxN];
vector<int> g[maxN];
int pr[maxN];
int deep[maxN];
void dfs(int v, int p = -1) {
  pr[v] = p;
  for (auto x : g[v]) {
    if (x == p) continue;
    deep[x] = deep[v] + 1;
    dfs(x, v);
    for (int k = 0; k < K; ++k) f[k][v] += max(1, f[k][x] - k);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  vector<pair<int, int>> vi(n);
  vector<int> dp(n);
  for (int i = 0; i < n; ++i) {
    vi[i].second = i;
    vi[i].first = g[i].size();
    if (i) --vi[i].first;
    dp[vi[i].second] = vi[i].first;
  }
  sort(vi.begin(), vi.end());
  int q;
  cin >> q;
  vector<int> used(n);
  while (q--) {
    int v, k;
    cin >> v >> k;
    --v;
    if (k < K) {
      cout << f[k][v] << "\n";
      continue;
    }
    int t =
        lower_bound(vi.begin(), vi.end(), make_pair(k + 1, -1)) - vi.begin();
    vector<int> touch;
    priority_queue<pair<int, int>> q;
    for (int i = t; i < n; ++i) {
      q.push({deep[vi[i].second], vi[i].second});
      used[vi[i].second] = 1;
      touch.push_back(vi[i].second);
    }
    while (!q.empty()) {
      int v = q.top().second;
      q.pop();
      if (pr[v] != -1) {
        --dp[pr[v]];
        dp[pr[v]] += dp[v] - k;
        touch.push_back(pr[v]);
        if (!used[pr[v]] && dp[pr[v]] > k) {
          used[pr[v]] = 1;
          q.push({deep[pr[v]], pr[v]});
        }
      }
    }
    cout << dp[v] << "\n";
    for (auto x : touch) {
      used[x] = 0;
      if (x == 0)
        dp[x] = g[x].size();
      else
        dp[x] = g[x].size() - 1;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  srand(time(0));
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,abm,mmx,tune=native")
using namespace std;
long long gcd(long long i, long long j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 1e9 + 100;
const int mod = 1000000007;
const long double eps = 1e-11, pi = acos(-1);
const long long maxN = 200010, maxT = 1010, A = 179, K = 200;
mt19937 mt_rand(time(0));
long long bp(long long et, long long b) {
  b %= mod - 1;
  et %= mod;
  long long res = 1;
  for (int i = 30; i >= 0; --i) {
    res = (res * res) % mod;
    if ((b & (1 << i)) != 0) res = (res * et) % mod;
  }
  return res;
}
void panic() {
  cout << "impossible\n";
  exit(0);
}
int f[K][maxN];
vector<int> g[maxN];
int pr[maxN];
int deep[maxN];
void dfs(int v, int p = -1) {
  pr[v] = p;
  for (auto x : g[v]) {
    if (x == p) continue;
    deep[x] = deep[v] + 1;
    dfs(x, v);
    for (int k = 0; k < K; ++k) f[k][v] += max(1, f[k][x] - k);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0);
  vector<pair<int, int>> vi(n);
  vector<int> dp(n);
  for (int i = 0; i < n; ++i) {
    vi[i].second = i;
    vi[i].first = g[i].size();
    if (i) --vi[i].first;
    dp[vi[i].second] = vi[i].first;
  }
  sort(vi.begin(), vi.end());
  int q;
  cin >> q;
  vector<int> used(n);
  while (q--) {
    int v, k;
    cin >> v >> k;
    --v;
    if (k < K) {
      cout << f[k][v] << "\n";
      continue;
    }
    int t =
        lower_bound(vi.begin(), vi.end(), make_pair(k + 1, -1)) - vi.begin();
    vector<int> touch;
    priority_queue<pair<int, int>> q;
    for (int i = t; i < n; ++i) {
      q.push({deep[vi[i].second], vi[i].second});
      used[vi[i].second] = 1;
      touch.push_back(vi[i].second);
    }
    while (!q.empty()) {
      int v = q.top().second;
      q.pop();
      if (pr[v] != -1) {
        --dp[pr[v]];
        dp[pr[v]] += dp[v] - k;
        touch.push_back(pr[v]);
        if (!used[pr[v]] && dp[pr[v]] > k) {
          used[pr[v]] = 1;
          q.push({deep[pr[v]], pr[v]});
        }
      }
    }
    cout << dp[v] << "\n";
    for (auto x : touch) {
      used[x] = 0;
      if (x == 0)
        dp[x] = g[x].size();
      else
        dp[x] = g[x].size() - 1;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  srand(time(0));
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```