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;
int N;
int sz[100001], ans[100001], L[100001];
vector<int> zero;
vector<int> g[100001];
vector<pair<int, int>> query[100001];
vector<int> *vec[100001];
string name[100001];
map<int, set<string>> cnt;
void subtree_size(int u, int p, int l) {
  sz[u] = 1;
  L[u] = l;
  for (int v : g[u]) {
    if (v != p) {
      subtree_size(v, u, l + 1);
      sz[u] += sz[v];
    }
  }
}
void dfs(int u, int p, bool keep) {
  int mx = -1, big_child = -1;
  for (auto &v : g[u]) {
    if (v != p && sz[v] > mx) {
      mx = sz[v], big_child = v;
    }
  }
  for (auto &v : g[u]) {
    if (v != p && v != big_child) {
      dfs(v, u, 0);
    }
  }
  if (big_child != -1) {
    dfs(big_child, u, 1);
    vec[u] = vec[big_child];
  } else {
    vec[u] = new vector<int>();
  }
  vec[u]->push_back(u);
  cnt[L[u]].insert(name[u]);
  for (auto &v : g[u]) {
    if (v != p && v != big_child) {
      for (auto &it : *vec[v]) {
        vec[u]->push_back(it);
        cnt[L[it]].insert(name[it]);
      }
    }
  }
  for (pair<int, int> q : query[u]) {
    ans[q.second] = cnt[L[u] + q.first].size();
  }
  if (keep == 0) {
    for (auto &it : *vec[u]) {
      cnt[L[it]].erase(name[it]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, u, p, i;
  cin >> n;
  N = n;
  for (i = 1; i <= n; i++) {
    cin >> name[i];
    cin >> p;
    if (p == 0)
      zero.push_back(i);
    else {
      g[p].push_back(i);
      g[i].push_back(p);
    }
  }
  for (int u : zero) subtree_size(u, -1, 0);
  cin >> m;
  for (i = 1; i <= m; i++) {
    cin >> u >> p;
    query[u].push_back({p, i});
  }
  for (int u : zero) dfs(u, -1, 0);
  for (i = 1; i <= m; i++) {
    cout << ans[i] << "\n";
  }
  return 0;
}
