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;
const int maxn = 100020;
const int maxx = 10000;
const int MOd = 1e9 + 7;
int a, ar[maxn], dad[maxn], ans[maxn];
vector<int> w[maxn];
vector<pair<int, int> > m[maxn];
map<string, int> mp;
string s[maxn];
map<int, int> dn[maxn];
set<pair<int, int> > dn2[maxn];
int dep[maxn];
void dfs(int n) {
  for (int i = 0; i < w[n].size(); i++) {
    int h = w[n][i];
    dfs(h);
    if (dn2[n].size() < dn2[h].size())
      swap(dn[n], dn[h]), swap(dn2[n], dn2[h]), swap(dep[n], dep[h]);
    for (set<pair<int, int> >::iterator it = dn2[h].begin(); it != dn2[h].end();
         it++) {
      pair<int, int> t =
          pair<int, int>(it->first + dep[h] - dep[n], it->second);
      if (dn2[n].find(t) == dn2[n].end()) {
        dn2[n].insert(t);
        dn[n][t.first]++;
      }
    }
  }
  for (int i = 0; i < m[n].size(); i++)
    ans[m[n][i].second] = dn[n][m[n][i].first - dep[n]];
  dn2[n].insert(pair<int, int>(-dep[n], ar[n]));
  dn[n][-dep[n]]++;
  dep[n]++;
}
int main() {
  scanf("%d", &a);
  for (int i = 1; i <= a; i++) {
    cin >> s[i];
    mp[s[i]] = 1;
    scanf("%d", &dad[i]);
    w[dad[i]].push_back(i);
  }
  int cnt = 0;
  for (map<string, int>::iterator it = mp.begin(); it != mp.end(); it++)
    it->second = ++cnt;
  for (int i = 1; i <= a; i++) ar[i] = mp[s[i]];
  int que;
  scanf("%d", &que);
  for (int i = 1, j, k; i <= que; i++) {
    scanf("%d %d", &j, &k);
    m[j].push_back(pair<int, int>(k, i));
  }
  for (int i = 1; i <= a; i++) sort(m[i].begin(), m[i].end());
  dfs(0);
  for (int i = 1; i <= que; i++) printf("%d\n", ans[i]);
  return 0;
}
