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;
using LL = long long;
int tm = 0;
void dfs(int v, int p, int h, const vector<vector<int> > &g,
         vector<vector<int> > &pr, vector<pair<int, int> > &times,
         vector<int> &hs) {
  times[v].first = tm;
  tm++;
  pr[v][0] = p;
  for (int i = 1; i < pr[v].size(); ++i) {
    pr[v][i] = pr[pr[v][i - 1]][i - 1];
  }
  hs[v] = h;
  for (auto to : g[v]) {
    if (to != p) {
      dfs(to, v, h + 1, g, pr, times, hs);
    }
  }
  times[v].second = tm;
  tm++;
}
int find_k(int v, int k, const vector<vector<int> > &pr) {
  if (k == 0) {
    return v;
  }
  int up = pr[v][0];
  int step = 1;
  for (int i = 1; i < pr[v].size(); ++i) {
    int new_step = step * 2;
    if (new_step <= k) {
      step = new_step;
      up = pr[v][i];
    }
  }
  return find_k(up, k - step, pr);
}
int main() {
  cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  cout.setf(std::ios_base::fixed);
  cout.precision(24);
  int n;
  cin >> n;
  n++;
  vector<vector<int> > g(n);
  vector<vector<int> > pr(n, vector<int>(20, 0));
  vector<pair<int, int> > times(n);
  vector<int> hs(n);
  for (int i = 1; i < n; ++i) {
    int a;
    cin >> a;
    g[a].push_back(i);
  }
  dfs(0, 0, 0, g, pr, times, hs);
  vector<vector<int> > v_by_h(n);
  for (int i = 0; i < n; ++i) {
    v_by_h[hs[i]].push_back(times[i].first);
  }
  for (int i = 0; i < n; ++i) {
    sort(v_by_h[i].begin(), v_by_h[i].end());
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    if (hs[a] <= b) {
      cout << 0 << " ";
      continue;
    }
    int parent = find_k(a, b, pr);
    int hp = hs[parent];
    int hch = hp + b;
    cout << (lower_bound(v_by_h[hch].begin(), v_by_h[hch].end(),
                         times[parent].second) -
             lower_bound(v_by_h[hch].begin(), v_by_h[hch].end(),
                         times[parent].first)) -
                1
         << " ";
  }
  return 0;
}
