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;
struct Solution {
  vector<vector<int> > g;
  vector<int> ans;
  vector<vector<pair<int, int> > > q;
  vector<vector<pair<int, int> > > qq;
  struct data {
    int h;
    int sz;
    int st;
    int ft;
    int ver;
  };
  vector<data> a;
  int time = 0;
  vector<int> w;
  void prepare_dfs(int v, int cur_h = 0) {
    w.push_back(v);
    a[v].h = cur_h;
    a[v].sz = 1;
    a[time].ver = v;
    a[v].st = time++;
    for (auto& to : g[v]) {
      prepare_dfs(to, cur_h + 1);
      a[v].sz += a[to].sz;
    }
    a[v].ft = time;
    w.pop_back();
    for (auto& x : q[v]) {
      int id = w.size() - x.first;
      if (id >= 0 && w[id] > 0) {
        qq[w[id]].push_back(x);
      }
    }
  }
  map<int, int> cnt;
  void apply(int v) { ++cnt[a[v].h]; }
  void revert(int v) { --cnt[a[v].h]; }
  void calc_ans(int v) {
    for (auto& x : qq[v]) {
      ans[x.second] = cnt[a[v].h + x.first] - 1;
    }
  }
  void dfs(int v, bool keep) {
    int big_to = -1;
    int mx = 0;
    for (auto& to : g[v]) {
      if (a[to].sz > mx) {
        mx = a[to].sz;
        big_to = to;
      }
    }
    for (auto& to : g[v]) {
      if (to != big_to) {
        dfs(to, false);
      }
    }
    if (big_to != -1) {
      dfs(big_to, true);
    }
    apply(v);
    for (auto& to : g[v]) {
      if (to == big_to) continue;
      for (int t = a[to].st; t < a[to].ft; ++t) {
        apply(a[t].ver);
      }
    }
    calc_ans(v);
    if (keep == 0) {
      for (int t = a[v].st; t < a[v].ft; ++t) {
        revert(a[t].ver);
      }
    }
  }
  void solve() {
    int n;
    cin >> n;
    ++n;
    g.resize(n);
    qq.resize(n);
    q.resize(n);
    a.resize(n);
    int s = 0;
    for (int i = 1; i < n; ++i) {
      int p;
      cin >> p;
      g[p].push_back(i);
    }
    int m;
    cin >> m;
    ans.resize(m);
    for (int i = 0; i < m; ++i) {
      int v, p;
      cin >> v >> p;
      q[v].push_back(make_pair(p, i));
    }
    prepare_dfs(s);
    dfs(s, true);
    for (auto& x : ans) {
      cout << x << " ";
    }
  }
};
int main(void) {
  ios_base::sync_with_stdio(false);
  Solution().solve();
  return 0;
}