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;
int Next[100004], V[100004], Head[100004], tot;
void Add_Edge(int u, int v) { Next[++tot] = Head[u], V[Head[u] = tot] = v; }
map<unsigned long long, int> Mp[100004];
unsigned long long C[100004];
int n, q, K[100004], Sz[100004], Son[100004], Dep[100004];
char S[100004];
vector<int> Q[100004];
int Ans[100004], DFN[100004], Ed[100004], D[100004], dfn;
void Get(int A) {
  Sz[A] = 1;
  Son[A] = -1;
  int B;
  for (int i = Head[A]; i; i = Next[i]) {
    Dep[B = V[i]] = Dep[A] + 1;
    Get(B);
    Sz[A] += Sz[B];
    if (Son[A] == -1 || Sz[B] > Sz[Son[A]]) Son[A] = B;
  }
}
void Del(int A) {
  for (int i = (DFN[A]), i_end_ = (Ed[A] + 1); i < i_end_; i++) {
    int B = D[i];
    map<unsigned long long, int>::iterator it = Mp[Dep[B]].find(C[B]);
    it->second--;
    if (!it->second) Mp[Dep[B]].erase(it);
  }
}
void Add(int A) {
  for (int i = (DFN[A]), i_end_ = (Ed[A] + 1); i < i_end_; i++) {
    int B = D[i];
    Mp[Dep[B]][C[B]]++;
  }
}
void DFS(int A) {
  DFN[A] = ++dfn;
  D[dfn] = A;
  int B;
  for (int i = Head[A]; i; i = Next[i])
    if ((B = V[i]) != Son[A]) DFS(B), Del(B);
  if (~Son[A]) DFS(Son[A]);
  for (int i = Head[A]; i; i = Next[i])
    if ((B = V[i]) != Son[A]) Add(B);
  Mp[Dep[A]][C[A]]++;
  for (int i = (0), i_end_ = (Q[A].size()); i < i_end_; i++) {
    int j = Q[A][i];
    if (Dep[A] + K[j] < 100004) Ans[j] = Mp[Dep[A] + K[j]].size();
  }
  Ed[A] = dfn;
}
int main() {
  scanf("%d", &n);
  for (int i = (1), i_end_ = (n + 1); i < i_end_; i++) {
    int f;
    scanf("%s%d", S, &f);
    int m = strlen(S);
    for (int j = (0), j_end_ = (m); j < j_end_; j++) C[i] = C[i] * 233 + S[j];
    Add_Edge(f, i);
  }
  scanf("%d", &q);
  for (int i = (0), i_end_ = (q); i < i_end_; i++) {
    int x;
    scanf("%d %d", &x, &K[i]);
    Q[x].push_back(i);
  }
  Get(0);
  DFS(0);
  for (int i = (0), i_end_ = (q); i < i_end_; i++) printf("%d\n", Ans[i]);
  return 0;
}
