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;
long long n, m, cr[100055], ans[100055];
vector<long long> vr[100055];
map<string, long long> mr;
unordered_map<long long, long long> dis[100055];
vector<pair<long long, long long> > qr[100055];
vector<long long> root;
void inc(long long), dec(long long), dfs(long long, long long),
    func(long long, long long);
int main() {
  string str;
  long long p, tt = 0;
  scanf("%lld", &n);
  for (int i = 1; i < n + 1; i++) {
    cin >> str >> p;
    vr[p].push_back(i);
    if (mr.find(str) != mr.end())
      cr[i] = mr[str];
    else {
      cr[i] = tt;
      mr[str] = tt++;
    }
    if (p == 0) root.push_back(i);
  }
  scanf("%lld", &m);
  for (int i = 0; i < m; i++) {
    long long idx, d;
    scanf("%lld %lld", &idx, &d);
    qr[idx].push_back(make_pair(d, i));
  }
  for (int i = 0; i < root.size(); i++) {
    dfs(root[i], 0);
    func(root[i], 0);
  }
  for (int i = 0; i < m; i++) printf("%lld\n", ans[i]);
}
long long sr[100055], inv[100055], st[100055], ft[100055], dr[100055], tim = 0;
void dfs(long long idx, long long d) {
  dr[idx] = d;
  inv[tim] = idx;
  st[idx] = tim++;
  sr[idx] = 1;
  for (int i = 0; i < vr[idx].size(); i++) {
    dfs(vr[idx][i], d + 1);
    sr[idx] += sr[vr[idx][i]];
  }
  ft[idx] = tim;
}
void func(long long idx, long long keep) {
  long long m = 0, h = -1;
  for (int i = 0; i < vr[idx].size(); i++) {
    if (sr[vr[idx][i]] > m) {
      m = sr[vr[idx][i]];
      h = vr[idx][i];
    }
  }
  for (int i = 0; i < vr[idx].size(); i++) {
    if (vr[idx][i] != h) func(vr[idx][i], 0);
  }
  if (h != -1) func(h, 1);
  for (int i = 0; i < vr[idx].size(); i++) {
    if (vr[idx][i] != h) {
      for (int j = st[vr[idx][i]]; j < ft[vr[idx][i]]; j++) {
        inc(j);
      }
    }
  }
  inc(st[idx]);
  for (int i = 0; i < qr[idx].size(); i++) {
    if (dr[idx] + qr[idx][i].first > 100000)
      ans[qr[idx][i].second] = 0;
    else
      ans[qr[idx][i].second] = dis[dr[idx] + qr[idx][i].first].size();
  }
  if (!keep) {
    for (int i = st[idx]; i < ft[idx]; i++) dec(i);
  }
}
void inc(long long i) {
  long long x = cr[inv[i]];
  long long d = dr[inv[i]];
  dis[d][x]++;
  ;
}
void dec(long long i) {
  long long x = cr[inv[i]];
  long long d = dr[inv[i]];
  dis[d][x]--;
  if (dis[d][x] == 0) dis[d].erase(x);
}
