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 static java.util.Arrays.*;

import java.io.*;
import java.lang.reflect.*;
import java.util.*;

public class E {

	final int MOD = (int)1e9 + 7;
	final double eps = 1e-12;
	final int INF = (int)1e9;
	
	List<Integer> [][] CH;
	boolean [] V;
	
	void expand(int m) {
		for (int i = 0; CH[i][m] != null; ++i)
			for (int p : CH[i][m]) {
				if (V[p])
					expand(p);
				if (CH[i][p] != null) {
					if (CH[i+1][m] == null)
						CH[i+1][m] = new ArrayList<Integer>(1);
					CH[i+1][m].addAll(CH[i][p]);
				}
			}
		V[m] = false;
	}
	
	@SuppressWarnings("unchecked")
	public E () {
		Map<String, Integer> NAM = new HashMap<String, Integer>();

		int N = sc.nextInt();
		int [] C = new int[N];
		CH = new List [20][N];
		V = new boolean [N];
		
		for (int i = 0; i < N; ++i) {
			String name = sc.next();
			if (!NAM.containsKey(name))
				NAM.put(name, NAM.size());
			C[i] = NAM.get(name);
			int p = sc.nextInt() - 1;
			if (p >= 0) {
				if (CH[0][p] == null)
					CH[0][p] = new ArrayList<Integer>(1);
				CH[0][p].add(i);
			}
		}
		
		NAM = null;
		
		fill(V, true);
		for (int i = 0; i < N; ++i)
			if (V[i])
				expand(i);
		
		Map<Long, Integer> H = new HashMap<Long, Integer>();
		
		int M = sc.nextInt();
		for (int i = 0; i < M; ++i) {
			int v = sc.nextInt() - 1;
			int k = sc.nextInt();
			long key = 1L * v * N + k;
			if (!H.containsKey(key)) {
				Set<Integer> res = new HashSet<Integer>(0);
				List<Integer> X = new ArrayList<Integer>(); X.add(v);
				for (int b = 20; b >= 0; --b) {
					if ((k & (1 << b)) != 0) {
						List<Integer> Y = new ArrayList<Integer>();				
						for (int p : X)
							if (CH[b][p] != null)
								Y.addAll(CH[b][p]);
						X = Y;
					}
				}
				for (int p : X)
					res.add(C[p]);
				H.put(key, res.size());
			}
			print(H.get(key));
		}
	}

	////////////////////////////////////////////////////////////////////////////////////
	
	static MyScanner sc;
	
	static class MyScanner {
		public String next() {
			newLine();
			return line[index++];
		}
		
		public char nextChar() {
			return next().charAt(0);
		}
				
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
		
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		
		public String nextLine() {
			line = null;
			return readLine();
		}
		
		public String [] nextStrings() {
			line = null;
			return readLine().split(" ");
		}
		
		public char [] nextChars() {
			return next().toCharArray();
		}

		public Integer [] nextInts() {
			String [] L = nextStrings();
			Integer [] res = new Integer [L.length];
			for (int i = 0; i < L.length; ++i)
				res[i] = Integer.parseInt(L[i]);
			return res;
		}	
		
		public Long [] nextLongs() {
			String [] L = nextStrings();
			Long [] res = new Long [L.length];
			for (int i = 0; i < L.length; ++i)
				res[i] = Long.parseLong(L[i]);
			return res;
		}

		public Double [] nextDoubles() {
			String [] L = nextStrings();
			Double [] res = new Double [L.length];
			for (int i = 0; i < L.length; ++i)
				res[i] = Double.parseDouble(L[i]);
			return res;
		}

		//////////////////////////////////////////////
		
		private boolean eol() {
			return index == line.length;
		}

		private String readLine() {
			try {
				return r.readLine();
			} catch (Exception e) {
				throw new Error(e);
			}
		}
		private final BufferedReader r;

		MyScanner () {
			this(new BufferedReader(new InputStreamReader(System.in)));
		}
		
		MyScanner(BufferedReader r) {
			try {
				this.r = r;
				while (!r.ready())
					Thread.sleep(1);
				start();
			} catch (Exception e) {
				throw new Error(e);
			}
		}
		
		private String [] line;
		private int index;

		private void newLine() {
			if (line == null || eol()) {
				line = readLine().split(" ");
				index = 0;
			}
		}		
	}
	
	static void print (Object... a) {
		pw.println(build(a));
	}

	static void exit (Object... a) {
		print(a);
		exit();
	}

	static void exit () {
		pw.close();
		System.out.flush();
		System.err.println("------------------");
		System.err.println("Time: " + ((millis() - t) / 1000.0));
		System.exit(0);
	}
	
	void NO() {
		throw new Error("NO!");
	}
	
	////////////////////////////////////////////////////////////////////////////////////
	
	static String build(Object... a) {
		StringBuilder b = new StringBuilder();
		for (Object o : a)
			append(b, o);
		return b.toString().trim();		
	}
	
	static void append(StringBuilder b, Object o) {
		if (o.getClass().isArray()) {
			int L = Array.getLength(o);
			for (int i = 0; i < L; ++i)
				append(b, Array.get(o, i));
		} else if (o instanceof Iterable<?>) {
			for (Object p : (Iterable<?>)o)
				append(b, p);
		} else
			b.append(" ").append(o);		
	}
	
	////////////////////////////////////////////////////////////////////////////////////
	
	public static void main(String[] args) {
		sc = new MyScanner ();
		new E();
		exit();
	}

	static void start() {
		t = millis();
	}

	static PrintWriter pw = new PrintWriter(System.out);
	
	static long t;
	
	static long millis() {
		return System.currentTimeMillis();
	}	
}