Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int MAXN = 1e5 + 5;
int N, M;
int A[MAXN], cntN;
std::vector<int> G[MAXN];
std::map<std::string, int> Str;
std::vector<std::pair<int, int>> Q[MAXN];
int Sz[MAXN], Son[MAXN], Dep[MAXN];
void getSon(int x) {
  Sz[x] = 1;
  for (auto v : G[x]) {
    Dep[v] = Dep[x] + 1;
    getSon(v);
    Sz[x] += Sz[v];
    if (Sz[v] > Sz[Son[x]]) {
      Son[x] = v;
    }
  }
}
int Ans[MAXN];
int skip;
std::multiset<int> S[MAXN];
int Cnt[MAXN];
void add(int x, int val) {
  if (val == 1) {
    if (S[Dep[x]].find(A[x]) == S[Dep[x]].end()) {
      ++Cnt[Dep[x]];
    }
    S[Dep[x]].insert(A[x]);
  } else {
    S[Dep[x]].erase(S[Dep[x]].find(A[x]));
    if (S[Dep[x]].find(A[x]) == S[Dep[x]].end()) {
      --Cnt[Dep[x]];
    }
  }
  for (auto v : G[x]) {
    if (v != skip) {
      add(v, val);
    }
  }
}
void dfs(int x, bool keep) {
  for (auto v : G[x]) {
    if (v != Son[x]) {
      dfs(v, 0);
    }
  }
  if (Son[x]) {
    dfs(Son[x], 1);
    skip = Son[x];
  }
  add(x, 1);
  for (auto v : Q[x]) {
    if (Dep[x] + v.first <= N) {
      Ans[v.second] = Cnt[Dep[x] + v.first];
    }
  }
  skip = 0;
  if (!keep) {
    add(x, -1);
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin >> N;
  int u, v;
  std::string name;
  for (int i = 1; i <= N; ++i) {
    std::cin >> name >> v;
    if (Str.find(name) == Str.end()) {
      Str[name] = ++cntN;
    }
    A[i] = Str[name];
    G[v].push_back(i);
  }
  std::cin >> M;
  for (int i = 1; i <= M; ++i) {
    std::cin >> u >> v;
    Q[u].push_back(std::make_pair(v, i));
  }
  getSon(0);
  dfs(0, 1);
  for (int i = 1; i <= M; ++i) {
    std::cout << Ans[i] << "\n";
  }
  return 0;
}