Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. 

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples

Input

6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1


Output

2
2
0
1
0


Input

6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

1
0
0
0
2
0
0
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const int M = 2e6 + 10;
int n, cnt, cntt, h[N], root[N], ans[N], d[N], q, val[N], cnttt, cntttt,
    num = 0;
set<int> s[M];
map<string, int> mpp;
struct segtree {
  int l, r, id;
} tree[M];
struct graph {
  int head[N], nxt[N << 1], to[N << 1], w[N << 1], sz;
  void init() { memset(head, 0, sizeof(head)), sz = 0; }
  graph() { init(); }
  void push(int a, int b, int c) {
    nxt[++sz] = head[a], to[sz] = b, w[sz] = c, head[a] = sz;
  }
  int &operator[](const int a) { return to[a]; }
} g;
struct Que {
  int k, id;
};
vector<Que> vec[100005];
void dfs(int x, int dep) {
  d[x] = dep;
  for (int i = g.head[x]; i; i = g.nxt[i]) {
    dfs(g[i], dep + 1);
  }
}
void insert(int p, int l, int r, int pos, int vl) {
  if (l == r) {
    if (!tree[p].id) tree[p].id = ++cntttt;
    s[tree[p].id].insert(vl);
    return;
  }
  int mid = l + r >> 1;
  if (pos <= mid) {
    if (!tree[p].l) tree[p].l = ++num;
    insert(tree[p].l, l, mid, pos, vl);
  } else {
    if (!tree[p].r) tree[p].r = ++num;
    insert(tree[p].r, mid + 1, r, pos, vl);
  }
}
int merge(int p, int q, int l, int r) {
  if (!p) return q;
  if (!q) return p;
  if (l == r) {
    if (!tree[p].id)
      tree[p].id = tree[q].id;
    else {
      if (s[tree[p].id].size() < s[tree[q].id].size())
        swap(tree[p].id, tree[q].id);
      set<int>::iterator it;
      for (it = s[tree[q].id].begin(); it != s[tree[q].id].end(); it++)
        s[tree[p].id].insert(*it);
    }
    return p;
  }
  int mid = l + r >> 1;
  tree[p].l = merge(tree[p].l, tree[q].l, l, mid);
  tree[p].r = merge(tree[p].r, tree[q].r, mid + 1, r);
  return p;
}
int query(int p, int l, int r, int x) {
  if (l == r) return s[tree[p].id].size();
  int mid = l + r >> 1;
  if (x <= mid)
    return query(tree[p].l, l, mid, x);
  else
    return query(tree[p].r, mid + 1, r, x);
}
void solve(int now) {
  insert(root[now], 1, n, d[now], val[now]);
  for (int i = g.head[now]; i; i = g.nxt[i]) {
    solve(g[i]);
    root[now] = merge(root[now], root[g[i]], 1, n);
  }
  for (Que y : vec[now]) ans[y.id] = query(root[now], 1, n, y.k);
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x; i <= n; i++) {
    root[i] = ++num;
    string str;
    cin >> str;
    if (!mpp[str])
      val[i] = mpp[str] = ++cnttt;
    else
      val[i] = mpp[str];
    scanf("%d", &x);
    g.push(x, i, 0);
  }
  dfs(0, 0);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int x, k;
    scanf("%d%d", &x, &k);
    if (d[x] + k <= n) vec[x].push_back({d[x] + k, i});
  }
  solve(0);
  for (int i = 1; i <= q; i++) printf("%d\n", ans[i]);
  return 0;
}
