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 maxn = 1000 * 100 + 10;
int n, q;
int par[maxn][40], h[maxn], s[maxn], f[maxn];
vector<int> lvl[maxn], adj[maxn];
int num = 0;
void dfs(int x) {
  s[x] = ++num;
  lvl[h[x]].push_back(s[x]);
  for (int i = 1; i < 32; i++) par[x][i] = par[par[x][i - 1]][i - 1];
  for (int i = 0; i < (int)adj[x].size(); i++) {
    int next = adj[x][i];
    h[next] = h[x] + 1;
    dfs(next);
  }
  f[x] = ++num;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    if (x != -1) {
      adj[x].push_back(i);
      par[i][0] = x;
    } else {
      par[i][0] = i;
      h[i] = 0;
    }
  }
  for (int i = 0; i < n; i++)
    if (par[i][0] == i) dfs(i);
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    int z = x;
    if (h[x] < y) {
      cout << 0 << endl;
      continue;
    }
    for (int j = 0; j < 30; j++)
      if ((y >> j) & 1) z = par[z][j];
    int q1 = lower_bound(lvl[y + h[z]].begin(), lvl[y + h[z]].end(), s[z]) -
             lvl[y + h[z]].begin();
    int q2 = lower_bound(lvl[y + h[z]].begin(), lvl[y + h[z]].end(), f[z]) -
             lvl[y + h[z]].begin();
    cout << q2 - q1 - 1 << endl;
  }
  return 0;
}
