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;
mt19937 rnd(0);
int id[200005], sub[200005], freq[200005], lv[200005];
unordered_multiset<int> occ[200005];
vector<int> adj[200005];
vector<pair<int, int>> ch[200005], query[200005];
int ans[200005];
void dfs(int x) {
  sub[x] = 1;
  for (auto it : adj[x]) {
    lv[it] = lv[x] + 1;
    dfs(it);
    sub[x] += sub[it];
  }
}
void dsu(int x) {
  int big = -1, mx = 0;
  for (auto it : adj[x]) {
    if (sub[it] > mx) mx = sub[it], big = it;
  }
  for (auto it : adj[x]) {
    if (it == big) continue;
    dsu(it);
    for (auto it2 : ch[it]) {
      auto it3 = occ[it2.first].find(it2.second);
      occ[it2.first].erase(it3);
      if (!occ[it2.first].count(it2.second)) freq[it2.first]--;
    }
  }
  if (big != -1) {
    dsu(big);
    swap(ch[x], ch[big]);
  }
  for (auto it : adj[x]) {
    if (it == big) continue;
    for (auto it2 : ch[it]) {
      if (!occ[it2.first].count(it2.second)) freq[it2.first]++;
      occ[it2.first].insert(it2.second);
      ch[x].push_back(it2);
    }
  }
  if (!occ[lv[x]].count(id[x])) freq[lv[x]]++;
  occ[lv[x]].insert(id[x]);
  ch[x].push_back({lv[x], id[x]});
  for (auto it : query[x]) {
    ans[it.first] = freq[lv[x] + it.second];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  int p;
  int cnt = 0;
  map<string, int> dict;
  string name;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> name >> p;
    if (dict.count(name))
      id[i] = dict[name];
    else {
      cnt++;
      id[i] = dict[name] = cnt;
    }
    adj[p].push_back(i);
  }
  int v, k, q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> v >> k;
    query[v].push_back({i, k});
  }
  lv[0] = 0;
  dfs(0);
  dsu(0);
  for (int i = 1; i <= q; i++) cout << ans[i] << '\n';
  return 0;
}
