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 = 2e5 + 5;
int n, q, ts, P[N], H[N], K[N], R[N], sz[N], F[N];
string S[N];
vector<int> Adj[N], Q[N];
map<string, int> MP;
map<int, int> M[N];
void Predfs(int v) {
  sz[v] = 1;
  for (int &u : Adj[v]) H[u] = H[v] + 1, Predfs(u), sz[v] += sz[u];
  for (int i = 0; i < Adj[v].size(); i++)
    if (sz[Adj[v][i]] > sz[Adj[v][0]]) swap(Adj[v][i], Adj[v][0]);
}
void Add(int v, int w) {
  M[H[v]][F[v]] += w;
  if (w == -1 && !M[H[v]][F[v]]) M[H[v]].erase(F[v]);
  for (int &u : Adj[v]) Add(u, w);
}
void DFS(int v) {
  M[H[v]][F[v]]++;
  if (!Adj[v].size()) return;
  for (int &u : Adj[v])
    if (u != Adj[v][0]) DFS(u), Add(u, -1);
  DFS(Adj[v][0]);
  for (int &u : Adj[v])
    if (u != Adj[v][0]) Add(u, 1);
  for (int &id : Q[v]) R[id] = M[K[id] + H[v]].size();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> S[i] >> P[i];
    Adj[P[i]].push_back(i);
    if (!MP[S[i]]) MP[S[i]] = ++ts;
    F[i] = MP[S[i]];
  }
  cin >> q;
  for (int i = 1, v; i <= q; i++) cin >> v >> K[i], Q[v].push_back(i);
  Predfs(0);
  DFS(0);
  for (int i = 1; i <= q; i++) cout << R[i] << '\n';
  return (0);
}
