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;
long long n, q;
vector<vector<long long> > a(100005);
long long col[100005];
map<string, long long> enc;
long long peak = 0;
vector<pair<long long, long long> > query[100005];
long long sta[100005], en[100005], child[100005], ver[100005], Time = 0,
                                                               h[100005];
map<long long, long long> cur[100005];
long long ansQuery[100005];
void pdfs(long long u, long long p) {
  sta[u] = ++Time;
  ver[Time] = u;
  child[u] = 1;
  for (auto v : a[u]) {
    if (v == p) continue;
    h[v] = h[u] + 1;
    pdfs(v, u);
    child[u] += child[v];
  }
  en[u] = Time;
}
void dfs(long long u, long long p, bool keep) {
  pair<long long, long long> mx = make_pair(-1, -1);
  for (auto v : a[u]) {
    if (v == p) continue;
    mx = max(mx, make_pair(child[v], v));
  }
  for (auto v : a[u]) {
    if (v == p || v == mx.second) continue;
    dfs(v, u, 0);
  }
  if (mx.second != -1) dfs(mx.second, u, 1);
  for (auto v : a[u]) {
    if (v == p || v == mx.second) continue;
    for (long long i = sta[v]; i <= en[v]; i++) cur[h[ver[i]]][col[ver[i]]]++;
    ;
  }
  cur[h[u]][col[u]]++;
  ;
  for (auto it : query[u]) {
    ansQuery[it.second] = cur[it.first + h[u]].size();
  }
  if (!keep) {
    for (long long i = sta[u]; i <= en[u]; i++) {
      cur[h[ver[i]]][col[ver[i]]]--;
      if (cur[h[ver[i]]][col[ver[i]]] == 0) cur[h[ver[i]]].erase(col[ver[i]]);
    };
  }
}
vector<long long> allRoot;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    string s;
    long long p;
    cin >> s >> p;
    if (enc[s] == 0) enc[s] = ++peak;
    if (p == 0) {
      allRoot.push_back(i);
      continue;
    }
    a[p].push_back(i);
    col[i] = enc[s];
  }
  cin >> q;
  for (long long i = 1; i <= q; i++) {
    long long u, k;
    cin >> u >> k;
    query[u].push_back(make_pair(k, i));
  }
  for (auto it : allRoot) pdfs(it, it), dfs(it, it, 0);
  for (long long i = 1; i <= q; i++) cout << ansQuery[i] << '\n';
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << '\n';
}
