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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
const int MAXN = 100005;
int n, m;
int vis[MAXN], p[MAXN], ans[MAXN];
string str[MAXN];
vector<int> adj[MAXN];
vector<pair<int, int>> queries[MAXN];
map<int, set<int>> st[MAXN];
map<string, int> mp;
int cnt[MAXN];
void merge(int a, int b) {
  if (st[a].size() < st[b].size()) st[a].swap(st[b]);
  for (auto it : st[b]) {
    if (st[a][it.first].size() < it.second.size()) {
      swap(st[a][it.first], it.second);
    }
    for (auto t : it.second) {
      st[a][it.first].insert(t);
    }
  }
}
void dfs(int s, int p, int h) {
  st[s][h].insert(mp[str[s]]);
  for (auto t : adj[s]) {
    if (t != p) {
      dfs(t, s, h + 1);
      merge(s, t);
    }
  }
  for (auto t : queries[s]) {
    ans[t.second] = st[s][h + t.first].size();
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int idx = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> str[i] >> p[i];
    if (mp.find(str[i]) == mp.end()) {
      mp[str[i]] = idx++;
    }
    adj[p[i]].push_back(i);
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    queries[x].push_back({y, i});
  }
  for (int i = 1; i <= n; i++) {
    if (p[i] == 0) {
      dfs(i, 0, 0);
    }
  }
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
}
