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;
long long N, Q, parent[200005], v, k, sz[200005], tot[200005], ans[200005];
string name[200005];
unordered_map<string, long long> MP[200005];
vector<long long> Graph[200005];
vector<pair<long long, long long> > qry[200005];
long long pre(long long node = 0) {
  sz[node] = 1;
  for (auto i : Graph[node]) sz[node] += pre(i);
  return sz[node];
}
void add(long long node, long long lev, long long x) {
  MP[lev][name[node]] += x;
  if (MP[lev][name[node]] == 1 && x == 1) tot[lev]++;
  if (MP[lev][name[node]] == 0 && x == -1) tot[lev]--;
  for (auto i : Graph[node]) add(i, lev + 1, x);
}
void dfs(long long node, long long lev, bool keep) {
  long long maxim = 0, bigChild = -1;
  for (auto i : Graph[node])
    if (sz[i] > maxim) maxim = sz[i], bigChild = i;
  for (auto i : Graph[node])
    if (i != bigChild) dfs(i, lev + 1, 0);
  if (bigChild != -1) dfs(bigChild, lev + 1, 1);
  for (auto i : Graph[node])
    if (i != bigChild) add(i, lev + 1, 1);
  MP[lev][name[node]]++;
  if (MP[lev][name[node]] == 1) tot[lev]++;
  for (auto i : qry[node])
    if (lev + i.first <= N) ans[i.second] = tot[lev + i.first];
  if (!keep) add(node, lev, -1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (long long i = 1; i <= N; i++) {
    cin >> name[i] >> parent[i];
    Graph[parent[i]].push_back(i);
  }
  cin >> Q;
  for (long long i = 0; i < Q; i++) {
    cin >> v >> k;
    qry[v].push_back({k, i});
  }
  pre();
  dfs(0, 0, 0);
  for (long long i = 0; i < Q; i++) cout << ans[i] << "\n";
  return 0;
}