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;
const int N = 2 * 1e5 + 100, M = 450;
int n, q, dp[N][M], B, ans[N], sz[N], f[M];
vector<int> e[N];
struct node {
  int v, k;
} sh[N];
inline void ckmin(int &a, int b) { a = ((a < b) ? a : b); }
inline void ckmax(int &a, int b) { a = ((a > b) ? a : b); }
inline int read() {
  int f = 1, x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  return x * f;
}
void dfs(int x, int fa) {
  for (int u : e[x])
    if (u != fa) {
      dfs(u, x);
      for (int k = 0; k <= B; k++) dp[x][k] += max(1, dp[u][k] - k);
    }
}
void dfs1(int x, int fa) {
  memset(dp[x], 0, sizeof(dp[x]));
  sz[x] = 1;
  for (int u : e[x])
    if (u != fa) {
      dfs1(u, x);
      memset(f, 0, sizeof(f));
      int lim = min(B, sz[x]);
      for (int i = 0; i <= lim; i++) {
        ckmax(f[i], dp[x][i] + 1);
        int up = min(B, sz[u]);
        for (int j = 0; j <= up && i + j + 1 <= B; j++)
          ckmax(f[i + j + 1], dp[x][i] + dp[u][j]);
      }
      memcpy(dp[x], f, sizeof(f));
      sz[x] += sz[u];
    }
}
signed main() {
  n = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    e[u].push_back(v);
    e[v].push_back(u);
  }
  B = sqrt(n);
  dfs(1, 1);
  q = read();
  for (int i = 1; i <= q; i++) {
    sh[i] = (node){read(), read()};
    if (sh[i].k <= B) ans[i] = dp[sh[i].v][sh[i].k];
  }
  dfs1(1, 1);
  printf("OK\n");
  for (int i = 1; i <= q; i++)
    if (sh[i].k > B) {
      int MAX = 0;
      for (int j = 0; j <= B; j++) MAX = max(MAX, dp[sh[i].v][j] - j * sh[i].k);
      ans[i] = MAX;
    }
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
}
