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 = 100008;
map<string, int> f;
unordered_map<int, int> b[N];
vector<int> a[N];
vector<pair<int, int> > que[N];
int n, m, nam[N], tot, ans[N], dep[N], siz[N], son[N], st[N], ed[N], id[N],
    cot[N + N];
int dfs(int x) {
  id[st[x] = ++tot] = x;
  siz[x] = 1;
  int pp = 0;
  for (auto p : a[x]) {
    dep[p] = dep[x] + 1;
    dfs(p);
    siz[x] += siz[p];
    if (siz[p] > pp) {
      pp = siz[p];
      son[x] = p;
    }
  }
  ed[x] = tot;
  return 0;
}
int dfsx(int x, int keep) {
  for (auto p : a[x])
    if (p != son[x]) dfsx(p, 0);
  if (son[x]) dfsx(son[x], 1);
  if (x) {
    for (auto p : a[x])
      if (p != son[x])
        for (int q = st[p]; q <= ed[p]; q++) {
          if (1 == ++b[dep[id[q]]][nam[id[q]]]) cot[dep[id[q]]]++;
        }
    if (1 == ++b[dep[x]][nam[x]]) cot[dep[x]]++;
    for (auto p : que[x]) ans[p.second] = cot[p.first];
  }
  if (keep == 0)
    for (int p = st[x]; p <= ed[x]; p++) {
      if (0 == --b[dep[id[p]]][nam[id[p]]]) cot[dep[id[p]]]--;
    }
  return 0;
}
int main(void) {
  std::ios::sync_with_stdio(false);
  int i, k, p1, p2;
  string str;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> str >> k;
    a[k].push_back(i);
    if (f.find(str) == f.end())
      f[str] = nam[i] = ++tot;
    else
      nam[i] = f[str];
  }
  cin >> m;
  tot = 0;
  dep[0] = 0;
  dfs(0);
  for (i = 1; i <= m; i++) {
    cin >> p1 >> p2;
    que[p1].push_back(make_pair(dep[p1] + p2, i));
  }
  dfsx(0, 0);
  for (i = 1; i <= m; i++) cout << ans[i] << endl;
  return 0;
}
