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>
#pragma comment(linker, "/STACK:16000000")
using namespace std;
const int Maxn = 100005;
const int Maxm = 20;
const int Maxt = 524288;
int n;
int p[Maxn];
vector<int> neigh[Maxn];
int cur, lef[Maxn], rig[Maxn];
int L[Maxn], P[Maxn][Maxm];
map<int, int> lvl[Maxt];
int m;
void Insert(int v, int l, int r, int pos, int val) {
  lvl[v][val]++;
  if (l < r) {
    int m = l + r >> 1;
    if (pos <= m)
      Insert(2 * v, l, m, pos, val);
    else
      Insert(2 * v + 1, m + 1, r, pos, val);
  }
}
int Get(int v, int l, int r, int a, int b, int val) {
  if (l == a && r == b)
    return lvl[v].count(val) ? lvl[v][val] : 0;
  else {
    int res = 0;
    int m = l + r >> 1;
    if (a <= m) res += Get(2 * v, l, m, a, min(m, b), val);
    if (m + 1 <= b) res += Get(2 * v + 1, m + 1, r, max(m + 1, a), b, val);
    return res;
  }
}
void Traverse(int v, int lvl) {
  lef[v] = cur++;
  P[v][0] = p[v];
  L[v] = lvl;
  Insert(1, 0, n - 1, lef[v], L[v]);
  for (int i = 0; i < neigh[v].size(); i++) Traverse(neigh[v][i], lvl + 1);
  rig[v] = cur - 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    if (p[i]) neigh[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++)
    if (!p[i]) Traverse(i, 0);
  for (int j = 1; j < Maxm; j++)
    for (int i = 1; i <= n; i++) P[i][j] = P[P[i][j - 1]][j - 1];
  scanf("%d", &m);
  bool was = false;
  while (m--) {
    int v, p;
    scanf("%d %d", &v, &p);
    int ans;
    if (p > L[v])
      ans = 0;
    else {
      int x = v;
      for (int i = Maxm - 1; i >= 0; i--)
        if (1 << i <= p) {
          x = P[x][i];
          p -= 1 << i;
        }
      ans = Get(1, 0, n - 1, lef[x], rig[x], L[v]) - 1;
    }
    if (was)
      printf(" ");
    else
      was = true;
    printf("%d", ans);
  }
  printf("\n");
  return 0;
}
