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;
inline long long labs(long long a) { return a < 0 ? (-a) : a; }
template <typename T>
inline T sqr(T x) {
  return x * x;
}
struct Node {
  int deep;
  int p;
  vector<int> to;
  int tin, tout;
};
const int MAXN = 100005;
vector<pair<int, int> > mas[MAXN];
vector<Node> g;
int n;
int t = 0;
void DFS(int i, int deep = 0) {
  g[i].deep = deep;
  g[i].tin = t++;
  mas[deep].push_back(make_pair(g[i].tin, i));
  for (int(j) = 0; (j) < (g[i].to.size()); (j)++) {
    int to = g[i].to[j];
    DFS(to, deep + 1);
  }
  g[i].tout = t++;
}
int main() {
  cin >> n;
  n++;
  g.resize(n);
  for (int(i) = 0; (i) < (n - 1); (i)++) {
    cin >> g[i + 1].p;
    g[g[i + 1].p].to.push_back(i + 1);
  }
  DFS(0);
  int m;
  cin >> m;
  for (int(qqq) = 0; (qqq) < (m); (qqq)++) {
    int i, ppp;
    cin >> i >> ppp;
    int pd = g[i].deep - ppp;
    if (pd <= 0) {
      cout << 0 << " ";
      continue;
    }
    int pit = upper_bound((mas[pd]).begin(), (mas[pd]).end(),
                          make_pair(g[i].tin, 0)) -
              mas[pd].begin() - 1;
    int p = mas[pd][pit].second;
    int ubl = upper_bound((mas[g[i].deep]).begin(), (mas[g[i].deep]).end(),
                          make_pair(g[p].tin, 0)) -
              mas[g[i].deep].begin();
    int ubr = upper_bound((mas[g[i].deep]).begin(), (mas[g[i].deep]).end(),
                          make_pair(g[p].tout, 0)) -
              mas[g[i].deep].begin() - 1;
    cout << ubr - ubl << " ";
  }
  cout << endl;
  return 0;
}
