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;
const int INFI = (1 << 30);
const long long INFL = (1LL << 62);
const double eps = 1e-7;
const long double pi = acos(-1.0);
const int MAXN = 100001;
int parent[MAXN], nrsons[MAXN];
vector<int> dson[MAXN];
vector<pair<int, int> > queries[MAXN], sol;
map<string, int> keys;
deque<set<int> > anc[MAXN];
string names[MAXN];
bool cmp(const pair<int, int>& p1, const pair<int, int>& p2) {
  return p1.second < p2.second;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int ancestor;
    cin >> names[i] >> ancestor;
    if (keys[names[i]] == 0) keys[names[i]] = keys.size();
    dson[ancestor].push_back(i);
    nrsons[ancestor]++;
    parent[i] = ancestor;
  }
  int Q;
  cin >> Q;
  for (int i = 1; i <= Q; i++) {
    int x, y;
    cin >> x >> y;
    queries[x].push_back(make_pair(y, i));
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (!dson[i].size()) {
      q.push(i);
    }
  }
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    if (dson[i].size()) {
      int mx = 0, mxi = 0;
      for (unsigned int j = 0; j < dson[i].size(); j++) {
        int u = dson[i][j];
        if (anc[u].size() > mx) {
          mx = anc[u].size();
          mxi = u;
        }
      }
      swap(anc[i], anc[mxi]);
    }
    for (unsigned int j = 0; j < dson[i].size(); j++) {
      int u = dson[i][j];
      for (unsigned int t = 0; t < anc[u].size(); t++) {
        if (anc[u][t].size() > anc[i][t].size()) {
          swap(anc[u][t], anc[i][t]);
        }
        anc[i][t].insert(anc[u][t].begin(), anc[u][t].end());
      }
      anc[u].clear();
    }
    set<int> ax;
    ax.insert(keys[names[i]]);
    anc[i].push_front(ax);
    for (unsigned int j = 0; j < queries[i].size(); j++) {
      if (queries[i][j].first < anc[i].size()) {
        sol.push_back(make_pair(anc[i][queries[i][j].first].size(),
                                queries[i][j].second));
      } else {
        sol.push_back(make_pair(0, queries[i][j].second));
      }
    }
    nrsons[parent[i]]--;
    if (!nrsons[parent[i]]) {
      q.push(parent[i]);
    }
  }
  sort(sol.begin(), sol.end(), cmp);
  for (unsigned int j = 0; j < sol.size(); j++) cout << sol[j].first << '\n';
  return 0;
}