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;
static vector<vector<int> > vc, vc2, vc1;
static int tin[1000000], tout[1000000], ti, le[1000000], pr[1000000],
    bu[20][200000];
void dfs(int n, int h) {
  le[n] = h;
  vc1[h].push_back(ti);
  tin[n] = ti++;
  bu[0][n] = pr[n];
  for (int i = 1; i < 19; ++i) bu[i][n] = bu[i - 1][bu[i - 1][n]];
  for (int i = 0; i < vc[n].size(); ++i) dfs(vc[n][i], h + 1);
  tout[n] = ti++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, a, b;
  cin >> n;
  string str;
  vc.resize(n + 1);
  vc1.resize(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> pr[i];
    if (pr[i])
      vc[pr[i]].push_back(i);
    else
      pr[i] = i;
  }
  for (int i = 1; i <= n; ++i)
    if (pr[i] == i) dfs(i, 0);
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> a >> b;
    if (le[a] < b) {
      cout << 0 << endl;
      continue;
    }
    for (int j = 0; j < 19; ++j)
      if ((1 << j) & b) a = bu[j][a];
    a;
    if (le[a] + b > n)
      cout << 0 << endl;
    else
      cout << upper_bound(vc1[le[a] + b].begin(), vc1[le[a] + b].end(),
                          tout[a]) -
                  upper_bound(vc1[le[a] + b].begin(), vc1[le[a] + b].end(),
                              tin[a]) -
                  1
           << '\n';
  }
  return 0;
}
