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;
int n, m, sz[100001], h[100001], ans[100001];
string s[100001];
vector<int> a[100001], vec[100001];
vector<pair<int, int>> query[100001];
map<string, int> son[100001];
void get_sz(int u, int p) {
  sz[u] = 1;
  for (auto v : a[u])
    if (v != p) {
      h[v] = h[u] + 1;
      get_sz(v, u);
      sz[u] += sz[v];
    }
}
void dfs(int u, int p, bool keep) {
  int big_sz = -1, big_child = -1;
  for (auto v : a[u])
    if (v != p && big_sz < sz[v]) {
      big_sz = sz[v];
      big_child = v;
    }
  for (auto v : a[u])
    if (v != p && v != big_child) dfs(v, u, false);
  if (big_child != -1) {
    dfs(big_child, u, true);
    swap(vec[u], vec[big_child]);
  }
  vec[u].push_back(u);
  son[h[u]][s[u]]++;
  for (auto v : a[u])
    if (v != p && v != big_child)
      for (auto x : vec[v]) {
        vec[u].push_back(x);
        son[h[x]][s[x]]++;
      }
  for (auto q : query[u])
    if (h[u] + q.second <= n)
      ans[q.first] = son[h[u] + q.second].size();
    else
      ans[q.first] = 0;
  if (!keep)
    for (auto v : vec[u]) {
      if (son[h[v]][s[v]] > 1)
        son[h[v]][s[v]]--;
      else
        son[h[v]].erase(s[v]);
    }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int r;
    cin >> s[i] >> r;
    a[r].push_back(i);
    a[i].push_back(r);
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int v, k;
    cin >> v >> k;
    query[v].push_back({i, k});
  }
  get_sz(0, -1);
  dfs(0, -1, true);
  for (int i = 1; i <= m; i++) cout << ans[i] << '\n';
  return 0;
}