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 MAX_N = 100 * 1000 + 10;
int n, q, name[MAX_N], q_k[MAX_N], ans[MAX_N];
vector<int> adj[MAX_N], roots, q_list[MAX_N];
unordered_map<int, unordered_set<int>> mp[MAX_N];
void read_jangle() {
  cin >> n;
  int cnt = 1;
  unordered_map<string, int> cnv;
  for (int i = 0; i < n; i++) {
    int par;
    string s;
    cin >> s >> par;
    if (cnv[s])
      name[i] = cnv[s];
    else
      name[i] = cnv[s] = cnt++;
    par--;
    if (par == -1)
      roots.push_back(i);
    else
      adj[par].push_back(i);
  }
}
void read_queries() {
  cin >> q;
  for (int i = 0; i < q; i++) {
    int v;
    cin >> v >> q_k[i];
    v--;
    q_list[v].push_back(i);
  }
}
void read_input() {
  read_jangle();
  read_queries();
}
void dfs(int, int);
void go_to_childs(int v, int cur_h) {
  mp[v][cur_h].insert(name[v]);
  for (int u : adj[v]) {
    dfs(u, cur_h + 1);
    if (mp[u].size() > mp[v].size()) swap(mp[u], mp[v]);
  }
}
void update_mp(int v) {
  for (int u : adj[v])
    for (auto i : mp[u]) {
      int h = i.first;
      for (int j : i.second) mp[v][h].insert(j);
    }
}
void fill_ans(int v, int cur_h) {
  for (int i : q_list[v]) {
    int h = q_k[i] + cur_h;
    ans[i] = mp[v][h].size();
  }
}
void dfs(int v, int cur_h = 0) {
  go_to_childs(v, cur_h);
  update_mp(v);
  fill_ans(v, cur_h);
}
void solve() {
  for (int r : roots) dfs(r);
}
void print_output() {
  for (int i = 0; i < q; i++) cout << ans[i] << endl;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  read_input();
  solve();
  print_output();
}
