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 JAVA):

Solution:
import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class E {

	int[][] par;
	ArrayList<Integer>[] a;
	ArrayList<Integer>[] len;
	int[] l;
	int[] d;
	int[] was;
	int[] dep;
	int time;

	void dfs(int v, int ds) {
		len[ds].add(v);
		time++;
		dep[v] = ds;
		d[v] = time;
		for (int i : a[v]) {
			dfs(i, ds + 1);
		}
		l[v] = time;
	}

	int findpar(int v, int len) {
		int res = v;
		int cur = 0;
		while (len > 0) {
			if (len % 2 == 1) {
				res = par[res][cur];
			}
			len /= 2;
			cur++;
		}
		return res;
	}

	int lower(int x, ArrayList<Integer> a) {
		int l = 0;
		int r = a.size() + 1;
		while (l < r - 1) {
			int mid = (l + r) / 2;
			if (d[a.get(mid - 1)] <= x) {
				l = mid;
			} else {
				r = mid;
			}
		}
		return l;
	}

	int find(int v, int deep) {
		deep += dep[v];
		return lower(l[v], len[deep]) - lower(d[v] - 1, len[deep]);
	}

	void count() throws IOException {
		int n = nextInt();
		par = new int[n + 1][20];
		l = new int[n + 1];
		d = new int[n + 1];
		dep = new int[n + 1];
		for (int i = 1; i <= n; i++) {
			par[i][0] = nextInt();
		}
		for (int j = 1; j < 20; j++) {
			for (int i = 1; i <= n; i++) {
				par[i][j] = par[par[i][j - 1]][j - 1];
			}
		}
		a = new ArrayList[n + 1];
		for (int i = 0; i <= n; i++) {
			a[i] = new ArrayList<Integer>();
		}
		len = new ArrayList[n + 1];
		for (int i = 0; i <= n; i++) {
			len[i] = new ArrayList<Integer>();
		}
		for (int i = 1; i <= n; i++) {
			a[par[i][0]].add(i);
		}
		dfs(0, 0);
		int m = nextInt();
		for (int i = 0; i < m; i++) {
			char last;
			if (i == m - 1) {
				last = '\n';
			} else {
				last = ' ';
			}
			int v = nextInt();
			int k = nextInt();
			if (k == 0) {
				throw new AssertionError();
			}
			int p1 = findpar(v, k);
			int p2 = findpar(v, k - 1);
			if (p1 * p2 == 0) {
				out.print(0 + "" + last);
			} else {
				out.print(find(p1, k) - 1 + "" + last);
			}
		}
	}

	BufferedReader in;
	PrintWriter out;
	StringTokenizer str;
	String readline;

	String nextToken() throws IOException {
		while (str == null || !str.hasMoreTokens()) {
			readline = in.readLine();
			if (readline == null) {
				return null;
			} else {
				str = new StringTokenizer(readline);
			}
		}
		return str.nextToken();
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	void run() throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		count();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new E().run();
	}
}