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 anc[19][100005], bt[100005], dt[100005], ht[100005], t;
int n, q, res[100005];
vector<int> v[100005], bh[100005];
struct qq {
  int node, lev, id;
};
vector<qq> query;
void dfs(int id, int pv) {
  if (pv != -1) {
    ht[id] = ht[pv] + 1;
    bt[id] = ++t;
    bh[ht[id]].push_back(bt[id]);
  }
  for (int i = (0); i <= ((int)v[id].size() - 1); i++) {
    int to = v[id][i];
    dfs(to, id);
  }
  if (pv != -1) dt[id] = ++t;
  return;
}
int search(int x, int dep) {
  int l = -1, r = bh[dep].size();
  while (r - l > 1) {
    int m = (l + r) >> 1;
    if (bh[dep][m] < x)
      l = m;
    else
      r = m;
  }
  return l + 1;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &anc[0][i]);
    v[anc[0][i]].push_back(i);
  }
  dfs(0, -1);
  for (int i = (1); i <= (17); i++)
    for (int j = (0); j <= (n); j++) anc[i][j] = anc[i - 1][anc[i - 1][j]];
  scanf("%d", &q);
  for (int i = (1); i <= (q); i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    if (ht[a] <= b)
      res[i] = 0;
    else {
      int ono = a;
      for (int j = (17); j >= (0); j--)
        if (b >= (1 << j)) {
          b -= 1 << j;
          ono = anc[j][ono];
        }
      res[i] = search(dt[ono], ht[a]) - search(bt[ono], ht[a]) - 1;
    }
    printf("%d%c", res[i], i == q ? '\n' : ' ');
  }
  return 0;
}
