246_E. Blood Cousins Return

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

Example 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

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


Example output:
2
2
0
1
0

1
0
0
0
2
0
0


Tags: binary search
data structures
dfs and similar
dp
sortings
Difficulty rating: 2400

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 200020;
vector<int> p[N], a[N];
map<pair<int, pair<int, int> >, int> g;
int c, n, m, x, y, d[N], l[N], r[N];
string s[N];
void fs(int x, int y) {
  p[y].push_back(x);
  l[x] = c++, d[x] = y;
  for (int &i : a[x]) fs(i, y + 1);
  r[x] = c;
}
int Q(int x, int y) {
  return lower_bound(p[x].begin(), p[x].end(), y,
                     [](int x, int y) { return l[x] < y; }) -
         p[x].begin();
}
int wk(int x, int y) {
  y += d[x];
  int L = Q(y, l[x]), R = Q(y, r[x]);
  set<string> q;
  int &t = g[make_pair(y, pair<int, int>(L, R))];
  if (t) return t;
  for (int i = L; i < R; i++) q.insert(s[p[y][i]]);
  return t = q.size();
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> s[i] >> x, a[x].push_back(i);
  fs(0, 0);
  for (cin >> m; m--;) cin >> x >> y, cout << wk(x, y) << endl;
}