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 N = 1e5 + 7;
int S[N], St[N], En[N], C[N], id[N], L[N], Ans[N], t;
vector<int> adj[N], root;
map<int, int> Map[4 * N];
vector<pair<int, int> > Query[N];
void dfs(int u, int p) {
  S[u] = 1;
  L[u] = L[p] + 1;
  St[u] = ++t;
  id[t] = u;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs(v, u);
    S[u] += S[v];
    if (S[v] > S[adj[u][0]]) swap(adj[u][i], adj[u][0]);
  }
  En[u] = t;
}
void dfs(int u, int p, int k) {
  for (int i = 1; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs(v, u, 0);
  }
  if (adj[u].size() && adj[u][0] != p) dfs(adj[u][0], u, 1);
  Map[L[u]][C[u]]++;
  for (int i = 1; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) continue;
    for (int j = St[v]; j <= En[v]; j++) {
      Map[L[id[j]]][C[id[j]]]++;
    }
  }
  for (auto it = Query[u].begin(); it != Query[u].end(); ++it) {
    Ans[it->first] = Map[L[u] + (it->second)].size();
  }
  if (!k) {
    for (int i = St[u]; i <= En[u]; i++) {
      int v = id[i];
      Map[L[v]][C[v]]--;
      if (!Map[L[v]][C[v]]) Map[L[v]].erase(C[v]);
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, I = 0;
  cin >> n;
  map<string, int> Map;
  for (int i = 1; i <= n; i++) {
    string s;
    int p;
    cin >> s >> p;
    if (!Map[s]) Map[s] = ++I;
    C[i] = Map[s];
    adj[i].push_back(p);
    adj[p].push_back(i);
    if (!p) root.push_back(i);
  }
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int v, h;
    cin >> v >> h;
    Query[v].push_back({i, h});
  }
  for (auto it : root) dfs(it, 0);
  for (auto it : root) dfs(it, 0, 0);
  for (int i = 1; i <= q; i++) {
    cout << Ans[i] << " ";
  }
}