Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100 * 1000 + 14, MAXL = 20;
int r[MAXN], par[MAXL][MAXN], h[MAXN], starting_time[MAXN],
    finishing_time[MAXN], _time = 0;
vector<int> adj[MAXN], height[MAXN];
void dfs(int v, int parent = -1) {
  starting_time[v] = _time++;
  par[0][v] = parent;
  for (int i = 1; par[i - 1][v] != -1 && i < 20; i++)
    par[i][v] = par[i - 1][par[i - 1][v]];
  for (auto u : adj[v]) {
    h[u] = h[v] + 1;
    dfs(u, v);
    height[h[u]].push_back(starting_time[u]);
  }
  finishing_time[v] = _time++;
  return;
}
int get_par(int v, int h) {
  for (int i = 0; i < MAXL; i++)
    if ((h >> i) & 1) v = par[i][v];
  return v;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(par, -1, sizeof(par));
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> r[i];
    adj[r[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (r[i] == 0) {
      dfs(i);
      height[0].push_back(starting_time[i]);
    }
  }
  int m;
  cin >> m;
  while (m--) {
    int v, p;
    cin >> v >> p;
    int parent = get_par(v, p);
    if (parent == -1) {
      cout << 0 << ' ';
      continue;
    }
    int s = lower_bound(height[h[v]].begin(), height[h[v]].end(),
                        starting_time[parent]) -
            height[h[v]].begin();
    int f = upper_bound(height[h[v]].begin(), height[h[v]].end(),
                        finishing_time[parent]) -
            height[h[v]].begin() - 1;
    cout << f - s << ' ';
  }
  cout << '\n';
  return 0;
}