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;
const int N = 1e5 + 5, L = 18;
int p[N][L], h[N];
vector<int> roots, adj[N], son[N][L];
void dfs(int v, int par, int hei) {
  p[v][0] = par;
  h[v] = hei++;
  for (int i = 1; i < L; i++) p[v][i] = p[p[v][i - 1]][i - 1];
  for (auto u : adj[v]) {
    son[v][0].push_back(u);
    dfs(u, v, hei);
  }
  for (int i = 1; i < L; i++)
    for (auto x : son[v][i - 1])
      for (auto y : son[x][i - 1]) son[v][i].push_back(y);
  return;
}
int ipar(int v, int x) {
  for (int i = 0; i < L; i++)
    if ((1 << i) & x) v = p[v][i];
  return v;
}
int ison(int v, int x) {
  for (int i = 0; i < L; i++)
    if (x == (1 << i)) return son[v][i].size();
  int res = 0;
  for (int i = 1; i < L; i++)
    if ((1 << i) & x) {
      for (auto u : son[v][i]) res += ison(u, x - (1 << i));
      return res;
    }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (!x)
      roots.push_back(i + 1);
    else
      adj[x].push_back(i + 1);
  }
  for (auto u : roots) dfs(u, u, 0);
  int m;
  scanf("%d", &m);
  while (m--) {
    int v, x, pi;
    scanf("%d%d", &v, &x);
    pi = ipar(v, x);
    if (h[v] - h[pi] != x) {
      printf("0\n");
      continue;
    }
    printf("%d\n", ison(pi, x) - 1);
  }
}
