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 + 4, B = 17;
vector<int> ed[N], pts, pin[N];
int n, m, par[B][N], h[N], ts[N], te[N], t[N], ct, tp[N];
void dfs(int x) {
  ts[x] = t[x] = ++ct;
  tp[t[x]] = x;
  pts.push_back(x);
  pin[h[x]].push_back(t[x]);
  for (int to : ed[x]) {
    h[to] = h[x] + 1;
    dfs(to);
  }
  te[x] = ct;
}
void process() {
  for (int b = 1; b < B; b++)
    for (int x : pts)
      par[b][x] = (par[b - 1][x] == -1) ? -1 : par[b - 1][par[b - 1][x]];
}
int solve(int x, int p) {
  if (h[x] < p) return 0;
  int oh = h[x];
  for (int b = B - 1; b >= 0; b--) {
    if (p >= (1 << b)) {
      x = par[b][x];
      p -= (1 << b);
    }
  }
  int st = lower_bound(pin[oh].begin(), pin[oh].end(), ts[x]) - pin[oh].begin();
  int et = upper_bound(pin[oh].begin(), pin[oh].end(), te[x]) - pin[oh].begin();
  return et - st - 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", par[0] + i);
    if (par[0][i])
      ed[par[0][i]].push_back(i);
    else
      par[0][i] = -1;
  }
  for (int i = 1; i <= n; i++) {
    if (par[0][i] == -1) {
      dfs(i);
      process();
      pts.clear();
    }
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int x, p;
    scanf("%d%d", &x, &p);
    printf("%d ", solve(x, p));
  }
  puts("");
}
