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>
using namespace std;
int N;
int id[110000], deep[110000], used[110000];
string name[110000];
map<string, int> mp;
map<pair<int, int>, int> sol;
vector<int> arb[110000], levels[110000];
int rad[110000];
int X, L[110000], R[110000];
void doBFS(int nod, int depth) {
  levels[depth].push_back(nod);
  deep[nod] = depth;
  L[nod] = ++X;
  for (int i = 0; i < arb[nod].size(); ++i) doBFS(arb[nod][i], depth + 1);
  R[nod] = ++X;
}
int binary_search(int depth, int elem) {
  int st = 0, dr = levels[depth].size();
  while (st < dr) {
    int m = (st + dr) / 2;
    if (L[levels[depth][m]] < elem)
      st = m + 1;
    else
      dr = m;
  }
  return st;
}
int query(int nod, int depth) {
  if (sol.find(make_pair(nod, depth)) != sol.end())
    return sol[make_pair(nod, depth)];
  int left = binary_search(depth, L[nod]);
  int right = binary_search(depth, R[nod]);
  int rez = 0;
  for (int i = left; i < right; ++i)
    if (!used[id[levels[depth][i]]]) {
      used[id[levels[depth][i]]] = 1;
      ++rez;
    }
  for (int i = left; i < right; ++i) used[id[levels[depth][i]]] = 0;
  sol[make_pair(nod, depth)] = rez;
  return rez;
}
void proc() {
  for (int i = 1; i <= N; ++i)
    if (rad[i]) doBFS(i, 0);
}
int main() {
  cin >> N;
  int pz = 0;
  for (int i = 1; i <= N; ++i) {
    cin >> name[i];
    if (mp.find(name[i]) == mp.end()) {
      ++pz;
      mp[name[i]] = pz;
    }
    id[i] = mp[name[i]];
    int nr;
    cin >> nr;
    if (nr != 0)
      arb[nr].push_back(i);
    else
      rad[i] = 1;
  }
  proc();
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; ++i) {
    int nod, depth;
    cin >> nod >> depth;
    cout << query(nod, deep[nod] + depth) << "\n";
  }
  return 0;
}