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;
#define lson (i<<1)
#define rson (i<<1|1)
#define repeat(i, s, n) for(int i = s; i <= n; ++i)
#define foreach(i,items) for(auto &i: items)
#define downrep(i, n, s) for(int i = n; i >= s; --i)
#define lowbit(x) (x&-x)
#define bzero(x) memset(x, 0, sizeof(x))
#define gcd(a,b) __gcd(a,b)

#define brkline puts("")
#define spaceout putchar(' ')
#define all(a) a.begin(),a.end()
#define pb  push_back

template<typename S, typename T>
bool chminx(S &a, const T &b){
    return a < b ? 1 : a=b, 0;
}

template<typename S, typename T>
bool chmaxx(S &a, const T &b){
    return a > b ? 1 : a=b, 0;
}

template<typename T>
void read_int(T &res) {
    int flag = 1;
    char ch;
    while(isspace(ch = getchar()));
    res = 0;
    if(ch == '-')  flag = -1;
    else if(ch >= '0' && ch <= '9')  res = ch - '0';
    while((ch=getchar()) >= '0' && ch <= '9')
        res = res * 10 + ch - '0';
    res *= flag;
}

template <typename T,typename... Args>
void read_ints(T &t,Args &...args){read_int(t);read_int(args...);}



template<typename T>
void int_out(T a) {
    if(a < 0) {
        putchar('-');
        a = -a;
    }
    if(a > 9)
        int_out(a / 10);
    putchar(a % 10 + '0');
}

typedef long long LL;

const int INF = 0x3f3f3f3f;
const double tol = 1e-8;
const int N = 200001;
const int SQ = sqrt(N);


int dp[N][SQ + 2], tmp[SQ + 2], sz[N];
// 开两个dp数组会超内存
int ks[N], vs[N], res[N];
vector<int> tree[N];

void get_dp1(int x, int fa){
    foreach(y, tree[x]) {
        if(y == fa) continue;
        get_dp1(y, x);
        repeat(k, 0, SQ) {
            dp[x][k] += max(dp[y][k] - k, 1);
        }
    }
}

void get_dp2(int x, int fa) {
    sz[x] = 1;
    foreach(y, tree[x]) {
        if(y == fa) continue;
        get_dp2(y, x);

        bzero(tmp);
        repeat(i, 0, min(sz[x], SQ)){
            chmaxx(tmp[i], dp[x][i] + 1);

            repeat(j, 0, sz[y]){
                if(i + j + 1 > SQ) break;
                chmaxx(tmp[i + j + 1], dp[x][i] + dp[x][j]);
            }
        }
        sz[x] += sz[y];
//        memcpy(dp[x], tmp, (min(sz[x], SQ) + 1) * sizeof *tmp);
        for(int i=0;i<=min(sz[x],SQ); ++i) dp[x][i]=tmp[i];
    }
}

int main(){

    int n;
    read_int(n);
    repeat(i, 1, n - 1) {
        int u, v;
        read_ints(u, v);
        tree[u].pb(v);
        tree[v].pb(u);
    }
    bzero(dp);
    get_dp1(1, 0);

    int q;
    read_int(q);
    bzero(res);
    repeat(i, 1, q) {
        read_ints(vs[i], ks[i]);
        if(ks[i] <= SQ) res[i] = dp[vs[i]][ks[i]];
    }

    bzero(dp);
    get_dp2(1, 0);
    repeat(i, 1, q) {
        int v = vs[i], k = ks[i];
        if(k > SQ) {
            repeat(j, 0, SQ) chmaxx(res[i], dp[v][j] - k * j);
        }
    }

    repeat(i, 1, q){
        int_out(res[i]);
        brkline;
    }

    return 0;
}
