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 N = 100100;
map<string, int> names[N];
int n, m, ssz[N], lvl[N], ans[N];
vector<int> tr[N];
vector<pair<int, int> > qrs[N];
string name[N];
void dfs0(int x = 0, int dpt = 0) {
  lvl[x] = dpt;
  ssz[x] = 1;
  for (auto& it : tr[x]) {
    dfs0(it, dpt + 1);
    ssz[x] += ssz[it];
  }
}
void add(int v) { names[lvl[v]][name[v]]++; }
void rem(int v) {
  auto& st = names[lvl[v]];
  st[name[v]]--;
  if (st[name[v]] == 0) st.erase(name[v]);
}
vector<int>* L[N];
void dfs(int x = 0, bool retain = 1) {
  int hvy = -1;
  for (auto& it : tr[x])
    if (hvy == -1 or ssz[it] > ssz[hvy]) hvy = it;
  for (auto& it : tr[x])
    if (it != hvy) dfs(it, 0);
  if (hvy != -1) {
    dfs(hvy, 1);
    L[x] = L[hvy];
  }
  if (L[x] == NULL) L[x] = new vector<int>();
  L[x]->emplace_back(x);
  add(x);
  for (auto& it : tr[x])
    if (it != hvy) {
      for (auto& jt : *L[it]) add(jt), L[x]->emplace_back(jt);
    }
  for (auto& qry : qrs[x]) {
    if (qry.first <= n) ans[qry.second] = names[qry.first].size();
  }
  if (!retain)
    for (auto& jt : *L[x]) rem(jt);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> name[i];
    int par;
    cin >> par;
    tr[par].emplace_back(i);
  }
  dfs0();
  cin >> m;
  for (int v, k, i = 1; i <= m; i++) {
    cin >> v >> k;
    qrs[v].emplace_back(lvl[v] + k, i);
  }
  dfs();
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}
