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 n;
vector<vector<int>> adj;
int raiz, reloj, l;
vector<int> tin, tout, altura;
vector<vector<int>> tinDeAltura, up;
void leerGrafo() {
  cin >> n;
  adj.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    int r;
    cin >> r;
    adj[r].push_back(i);
  }
}
void dfs(int p, int a, int h) {
  tin[a] = reloj++;
  altura[a] = h;
  tinDeAltura[altura[a]].push_back(tin[a]);
  up[a][0] = p;
  for (int i = 1; i <= l; ++i) up[a][i] = up[up[a][i - 1]][i - 1];
  for (auto e : adj[a]) dfs(a, e, h + 1);
  tout[a] = reloj++;
}
void precomputo() {
  raiz = 0;
  reloj = 0;
  l = ceil(log2(n));
  tin.resize(n + 1);
  tout.resize(n + 1);
  altura.resize(n + 1);
  up.resize(n + 1, vector<int>(l + 1));
  tinDeAltura.resize(n + 1);
  dfs(raiz, raiz, 0);
}
int sube(int a, int n) {
  int mask = 0;
  while (n) {
    if (n & 1) a = up[a][mask];
    n >>= 1;
    ++mask;
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  leerGrafo();
  precomputo();
  int m;
  cin >> m;
  while (m--) {
    int v, p;
    cin >> v >> p;
    if (p >= altura[v]) {
      cout << "0 ";
      continue;
    }
    v = sube(v, p);
    int ab = altura[v] + p;
    int L = (int)(upper_bound(tinDeAltura[ab].begin(), tinDeAltura[ab].end(),
                              tin[v]) -
                  tinDeAltura[ab].begin());
    int R = (int)(upper_bound(tinDeAltura[ab].begin(), tinDeAltura[ab].end(),
                              tout[v]) -
                  tinDeAltura[ab].begin());
    cout << R - L - 1 << " ";
  }
  cout << '\n';
  return 0;
}
