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;
void JIZZ() {
  cout << "";
  exit(0);
}
const long double PI = 3.14159265358979323846264338327950288;
const long double eps = 1e-13;
const long long mod = 1e9 + 7;
vector<int> G[100005];
int d[100005], ps[20][100005], in[100005], out[100005], timer;
vector<int> deps[100005];
void dfs(int now, int pa, int dep) {
  d[now] = dep;
  ps[0][now] = pa;
  in[now] = timer++;
  deps[dep].push_back(now);
  for (int i = 1; (1 << i) <= dep; ++i) ps[i][now] = ps[i - 1][ps[i - 1][now]];
  for (int i : G[now]) dfs(i, now, dep + 1);
  out[now] = timer++;
}
int main() {
  int n;
  cin >> n;
  queue<int> isRoot;
  for (int i = 1, pa; i <= n; ++i) {
    cin >> pa;
    if (pa == 0)
      isRoot.push(i);
    else
      G[pa].push_back(i);
  }
  while (isRoot.size()) {
    dfs(isRoot.front(), 0, 1);
    isRoot.pop();
  }
  int q;
  cin >> q;
  while (q--) {
    int u, v, p, dd;
    cin >> v >> p;
    u = v;
    for (int i = 0; i < 20; ++i)
      if ((1 << i) & p) u = ps[i][u];
    if (u == 0)
      cout << 0 << endl;
    else {
      auto it1 = lower_bound(deps[d[v]].begin(), deps[d[v]].end(), u,
                             [](const int &a, const int &b) {
                               ;
                               ;
                               return in[a] < in[b];
                             });
      auto it2 = lower_bound(deps[d[v]].begin(), deps[d[v]].end(), u,
                             [](const int &a, const int &b) {
                               ;
                               ;
                               return out[a] < out[b];
                             });
      ;
      ;
      cout << it2 - it1 - 1 << endl;
    }
  }
}
