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;
vector<vector<int>> conns;
vector<vector<pair<int, int>>> queries;
vector<vector<pair<int, int>>> resolvedQueries;
vector<int> results;
void resolveQueries(int pos, int prev, vector<int>& parents) {
  for (auto [dist, i] : queries[pos]) {
    if (dist <= parents.size()) {
      resolvedQueries[parents[parents.size() - dist]].emplace_back(dist, i);
    }
  }
  parents.push_back(pos);
  for (int p : conns[pos]) {
    if (p == prev) continue;
    resolveQueries(p, pos, parents);
  }
  parents.pop_back();
}
void fillResults(int pos, int prev, int depth, vector<int>& counts) {
  for (auto [dist, i] : resolvedQueries[pos]) {
    results[i] = -counts[depth + dist];
  }
  counts[depth]++;
  for (int p : conns[pos]) {
    if (p == prev) continue;
    fillResults(p, pos, depth + 1, counts);
  }
  for (auto [dist, i] : resolvedQueries[pos]) {
    results[i] += counts[depth + dist];
  }
}
int main() {
  int n, m;
  cin >> n;
  conns.resize(n);
  queries.resize(n);
  resolvedQueries.resize(n);
  vector<int> roots;
  for (int i = 0; i < n; i++) {
    int p;
    cin >> p;
    if (p > 0) {
      p--;
      conns[i].push_back(p);
      conns[p].push_back(i);
    } else {
      roots.push_back(i);
    }
  }
  cin >> m;
  results.resize(m);
  for (int i = 0; i < m; i++) {
    int pos, dist;
    cin >> pos >> dist;
    pos--;
    queries[pos].emplace_back(dist, i);
  }
  vector<int> counts(n);
  for (int r : roots) {
    vector<int> parents;
    resolveQueries(r, r, parents);
    fillResults(r, r, 0, counts);
  }
  for (int r : results) {
    cout << max(0, r - 1) << "\n";
  }
}
