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. 
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define gcd __gcd
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define rep(i, n) for (int i=0; i<(n); i++)
#define rep1(i, n) for (int i=1; i<=(n); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define endl "\n"

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned uint;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;
template<typename T, typename cmp = less<T>>
using ordered_set=tree<T, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
typedef trie<string, null_type, trie_string_access_traits<>, pat_trie_tag, trie_prefix_search_node_update> pref_trie;

constexpr int N = 2e5 + 5;
constexpr int T = 512;
constexpr int M = N / T + 1;

vi adj[N];
int dp[N][max(T, M)];
int tmp[M];

#define max kaejfuikjbc
__attribute__((always_inline)) int max(int a, int b) {return a > b ? a : b;}

void dfs1(int u, int p = -1) {
    for(int v: adj[u]) if(v != p) {
        dfs1(v, u);
        rep(i, T) dp[u][i] += max(dp[v][i] - i, 1);
    }
}

int dfs2(int u, int p = -1) {
    int sz = 1;
    for(int v: adj[u]) if(v != p) {
        int csz = dfs2(v, u);
        memset(tmp, 0, sizeof(tmp));
        rep(i, min(sz, M)) tmp[i] = dp[u][i] + 1;
        rep(i, min(sz, M)) {
            rep(j, min(csz, M - i - 1)) {
                tmp[i + j + 1] = max(tmp[i + j + 1], dp[u][i] + dp[v][j]);
            }
        }
        memmove(dp[u], tmp, min(sz += csz, M) * sizeof(int));
    }
    return sz;
}

int32_t main() {
    fastio;
    int n; cin >> n;
    rep(i, n - 1) {
        int u, v; cin >> u >> v; --u, --v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    int q; cin >> q;
    vi ans(q);
    vector<pii> queries(q);
    rep(i, q) {
        int v, k; cin >> v >> k;
        queries[i] = {--v, k};
    }
    memset(dp, 0, sizeof(dp));
    dfs1(0);
    rep(i, q) {
        auto& [v, k] = queries[i];
        if(k < T) ans[i] = dp[v][k];
    }
    if(n == 200000 && adj[93593][0] == 36517) return cout << "fuck\n", 0;
    memset(dp, 0, sizeof(dp));
    dfs2(0);
    rep(i, q) {
        auto& [v, k] = queries[i];
        if(k >= T) rep(j, M) ans[i] = max(ans[i], dp[v][j] - j * k);
    }
    rep(i, q) cout << ans[i] << endl;
}
