#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct node {
  int x, y;
  node(int x = 0, int y = 0) : x(x), y(y) {}
  bool operator<(const node c) const { return 1ll * y * c.x < 1ll * x * c.y; }
  bool operator>(const node c) const { return c < *this; }
  node operator+(const node c) const { return node(x + c.x, y + c.y); }
} val[200005], sum[200005];
mt19937 mrand(1);
int n, q, tot, ncnt;
int ch[200005][2], siz[200005], v[400005], nxt[400005], h[200005], son[200005],
    sz[200005], rt[200005], ans[200005];
vector<pair<int, int> > qry[200005];
void addedge(int x, int y) {
  v[++tot] = y;
  nxt[tot] = h[x];
  h[x] = tot;
  v[++tot] = x;
  nxt[tot] = h[y];
  h[y] = tot;
}
void dfs1(int u, int fa) {
  sz[u] = 1;
  for (int p = h[u]; p; p = nxt[p]) {
    if (v[p] == fa) continue;
    dfs1(v[p], u);
    sz[u] += sz[v[p]];
    if (sz[v[p]] > sz[son[u]]) son[u] = v[p];
  }
}
void update(int u) {
  siz[u] = 1 + siz[ch[u][0]] + siz[ch[u][1]];
  sum[u] = val[u] + sum[ch[u][0]] + sum[ch[u][1]];
}
void split(int u, node k, int &x, int &y) {
  if (!u) return (void)(x = y = 0);
  if (!(val[u] < k))
    x = u, split(ch[u][1], k, ch[x][1], y);
  else
    y = u, split(ch[u][0], k, x, ch[y][0]);
  update(u);
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  if (mrand() % (siz[x] + siz[y]) < siz[x]) {
    ch[x][1] = merge(ch[x][1], y);
    update(x);
    return x;
  } else {
    ch[y][0] = merge(x, ch[y][0]);
    update(y);
    return y;
  }
}
void dfs3(int u, int &t) {
  if (!u) return;
  int ta, tb;
  split(t, val[u], ta, tb);
  dfs3(ch[u][0], ta);
  dfs3(ch[u][1], tb);
  ch[u][0] = ch[u][1] = 0;
  update(u);
  t = merge(merge(ta, u), tb);
}
int getlf(int u) {
  while (ch[u][0]) u = ch[u][0];
  return u;
}
void dfs2(int u, int fa) {
  for (int p = h[u]; p; p = nxt[p])
    if (v[p] != fa && v[p] != son[u]) dfs2(v[p], u);
  int num = -1;
  if (son[u]) dfs2(son[u], u), rt[u] = rt[son[u]], num++;
  for (int p = h[u]; p; p = nxt[p]) {
    if (v[p] == fa || v[p] == son[u]) continue;
    dfs3(rt[v[p]], rt[u]);
    num++;
  }
  int a, b;
  for (auto r : qry[u]) {
    node tmp(1, r.first);
    split(rt[u], tmp, a, b);
    ans[r.second] = sum[a].y - 1ll * r.first * sum[a].x + num + 1;
    rt[u] = merge(a, b);
  }
  node th(1, num);
  while (siz[rt[u]] >= 1) {
    int tmp = getlf(rt[u]);
    if (th > val[tmp]) break;
    split(rt[u], val[tmp], a, b);
    th = th + val[a];
    rt[u] = merge(merge(ch[a][0], ch[a][1]), b);
  }
  ncnt++;
  val[ncnt] = sum[ncnt] = th, siz[ncnt] = 1;
  rt[u] = merge(ncnt, rt[u]);
}
int main() {
  n = readint();
  for (int i = 1; i < n; i++) addedge(readint(), readint());
  q = readint();
  for (int i = 1; i <= q; i++) {
    int x = readint(), y = readint();
    qry[x].push_back(make_pair(y, i));
  }
  dfs1(1, -1);
  dfs2(1, -1);
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, 1 : 0;
}
int readint() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct node {
  int x, y;
  node(int x = 0, int y = 0) : x(x), y(y) {}
  bool operator<(const node c) const { return 1ll * y * c.x < 1ll * x * c.y; }
  bool operator>(const node c) const { return c < *this; }
  node operator+(const node c) const { return node(x + c.x, y + c.y); }
} val[200005], sum[200005];
mt19937 mrand(1);
int n, q, tot, ncnt;
int ch[200005][2], siz[200005], v[400005], nxt[400005], h[200005], son[200005],
    sz[200005], rt[200005], ans[200005];
vector<pair<int, int> > qry[200005];
void addedge(int x, int y) {
  v[++tot] = y;
  nxt[tot] = h[x];
  h[x] = tot;
  v[++tot] = x;
  nxt[tot] = h[y];
  h[y] = tot;
}
void dfs1(int u, int fa) {
  sz[u] = 1;
  for (int p = h[u]; p; p = nxt[p]) {
    if (v[p] == fa) continue;
    dfs1(v[p], u);
    sz[u] += sz[v[p]];
    if (sz[v[p]] > sz[son[u]]) son[u] = v[p];
  }
}
void update(int u) {
  siz[u] = 1 + siz[ch[u][0]] + siz[ch[u][1]];
  sum[u] = val[u] + sum[ch[u][0]] + sum[ch[u][1]];
}
void split(int u, node k, int &x, int &y) {
  if (!u) return (void)(x = y = 0);
  if (!(val[u] < k))
    x = u, split(ch[u][1], k, ch[x][1], y);
  else
    y = u, split(ch[u][0], k, x, ch[y][0]);
  update(u);
}
int merge(int x, int y) {
  if (!x || !y) return x + y;
  if (mrand() % (siz[x] + siz[y]) < siz[x]) {
    ch[x][1] = merge(ch[x][1], y);
    update(x);
    return x;
  } else {
    ch[y][0] = merge(x, ch[y][0]);
    update(y);
    return y;
  }
}
void dfs3(int u, int &t) {
  if (!u) return;
  int ta, tb;
  split(t, val[u], ta, tb);
  dfs3(ch[u][0], ta);
  dfs3(ch[u][1], tb);
  ch[u][0] = ch[u][1] = 0;
  update(u);
  t = merge(merge(ta, u), tb);
}
int getlf(int u) {
  while (ch[u][0]) u = ch[u][0];
  return u;
}
void dfs2(int u, int fa) {
  for (int p = h[u]; p; p = nxt[p])
    if (v[p] != fa && v[p] != son[u]) dfs2(v[p], u);
  int num = -1;
  if (son[u]) dfs2(son[u], u), rt[u] = rt[son[u]], num++;
  for (int p = h[u]; p; p = nxt[p]) {
    if (v[p] == fa || v[p] == son[u]) continue;
    dfs3(rt[v[p]], rt[u]);
    num++;
  }
  int a, b;
  for (auto r : qry[u]) {
    node tmp(1, r.first);
    split(rt[u], tmp, a, b);
    ans[r.second] = sum[a].y - 1ll * r.first * sum[a].x + num + 1;
    rt[u] = merge(a, b);
  }
  node th(1, num);
  while (siz[rt[u]] >= 1) {
    int tmp = getlf(rt[u]);
    if (th > val[tmp]) break;
    split(rt[u], val[tmp], a, b);
    th = th + val[a];
    rt[u] = merge(merge(ch[a][0], ch[a][1]), b);
  }
  ncnt++;
  val[ncnt] = sum[ncnt] = th, siz[ncnt] = 1;
  rt[u] = merge(ncnt, rt[u]);
}
int main() {
  n = readint();
  for (int i = 1; i < n; i++) addedge(readint(), readint());
  q = readint();
  for (int i = 1; i <= q; i++) {
    int x = readint(), y = readint();
    qry[x].push_back(make_pair(y, i));
  }
  dfs1(1, -1);
  dfs2(1, -1);
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}

```