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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
}
template <typename T>
void print(T x) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
}
template <typename T>
void print(T x, char c) {
  print(x);
  putc(c);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MAXN = 2e5;
const int LOG_N = 18;
int n, qu, hd[MAXN + 5], to[MAXN * 2 + 5], nxt[MAXN * 2 + 5], ec = 0;
void adde(int u, int v) {
  to[++ec] = v;
  nxt[ec] = hd[u];
  hd[u] = ec;
}
int chd[MAXN + 5], dep[MAXN + 5], fa[MAXN + 5][LOG_N + 2];
int dfn[MAXN + 5], tim = 0, rid[MAXN + 5];
void dfs0(int x, int f) {
  fa[x][0] = f;
  rid[dfn[x] = ++tim] = x;
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f) continue;
    chd[x]++;
    dep[y] = dep[x] + 1;
    dfs0(y, x);
  }
}
int sum_chd[MAXN + 5];
void dfs_chd(int x, int f) {
  for (int e = hd[x]; e; e = nxt[e]) {
    int y = to[e];
    if (y == f) continue;
    sum_chd[y] = sum_chd[x] + chd[y];
    dfs_chd(y, x);
  }
}
void lca_init() {
  dep[1] = 1;
  dfs0(1, 0);
  for (int i = 1; i <= LOG_N; i++)
    for (int j = 1; j <= n; j++) fa[j][i] = fa[fa[j][i - 1]][i - 1];
}
int getlca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int i = LOG_N; ~i; i--)
    if (dep[x] - (1 << i) >= dep[y]) x = fa[x][i];
  if (x == y) return x;
  for (int i = LOG_N; ~i; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}
vector<int> pos[MAXN + 5];
int stk[MAXN + 5], tp = 0;
set<int> st;
vector<pair<int, int> > qv[MAXN + 5];
long long calc_sum(int u, int v, int k) {
  return (sum_chd[v] - sum_chd[u]) - 1ll * (k + 1) * (dep[v] - dep[u]);
}
int calc_val(int u, int k) { return chd[u] - 1 - k; }
vector<int> g[MAXN + 5];
long long dp[MAXN + 5], res[MAXN + 5];
void insert(int x) {
  if (!tp) return stk[++tp] = x, void();
  int lc = getlca(x, stk[tp]);
  while (tp > 1 && dep[stk[tp - 1]] > dep[lc]) {
    g[stk[tp - 1]].push_back(stk[tp]);
    --tp;
  }
  if (tp && dep[stk[tp]] > dep[lc]) {
    g[lc].push_back(stk[tp--]);
  }
  if (!tp || stk[tp] != lc) stk[++tp] = lc;
  stk[++tp] = x;
}
void fin() {
  while (tp >= 2) {
    g[stk[tp - 1]].push_back(stk[tp]);
    --tp;
  }
  tp = 0;
}
void dfs_dp(int x, int k) {
  dp[x] = calc_val(x, k);
  st.insert(dfn[x]);
  for (int y : g[x]) {
    dfs_dp(y, k);
    dp[x] += max(dp[y] + calc_sum(x, y, k) - calc_val(y, k), 0ll);
  }
}
void clr(int x) {
  for (int y : g[x]) clr(y);
  g[x].clear();
}
int main() {
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++)
    scanf("%d%d", &u, &v), adde(u, v), adde(v, u);
  lca_init();
  sum_chd[1] = chd[1];
  dfs_chd(1, 0);
  scanf("%d", &qu);
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < chd[i]; j++) pos[j].push_back(i);
  for (int i = 1, u, k; i <= qu; i++)
    scanf("%d%d", &u, &k), qv[k].push_back(make_pair(u, i));
  for (int i = 0; i <= MAXN; i++) {
    st.clear();
    if (!pos[i].empty()) {
      sort(pos[i].begin(), pos[i].end(),
           [&](int x, int y) { return dfn[x] < dfn[y]; });
      if (pos[i][0] != 1) insert(1);
      for (int x : pos[i]) insert(x);
      fin();
      dfs_dp(1, i);
    }
    for (pair<int, int> p : qv[i]) {
      int u = p.first, id = p.second;
      set<int>::iterator it = st.lower_bound(dfn[u]);
      if (it == st.end())
        res[id] = chd[u];
      else {
        int pt = rid[*it];
        res[id] = chd[u] + max(0ll, dp[pt] + calc_sum(fa[pt][0], fa[u][0], i) -
                                        calc_val(u, i));
      }
    }
    clr(1);
  }
  for (int i = 1; i <= qu; i++) printf("%lld\n", res[i]);
  return 0;
}
