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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;


public class Main {
	static int id = 0;
	static int[] min, max, H, M;
	static ArrayList<Integer>[] g, L; 
	public static void main(String[] args) {
		int[] log = new int[1 << 22];
		for(int i = 0; i < 22; i++)
			log[1 << i] = i;
		
		InputReader r = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int N = r.nextInt();
		
		int[] P = new int[N + 1];
		P[0] = -1;
		for(int i = 0; i < N; i++)
			P[i + 1] = r.nextInt();
		
		H = new int[N + 1];
		
		M = new int[N + 1];
		
		int[][] U = new int[N + 1][22];
		for(int i = 0; i < N + 1; i++)
			U[i][0] = P[i];
		for(int h = 1; h < 22; h++)
		for(int i = 0; i < N + 1; i++)
			if(U[i][h - 1] != -1)U[i][h] = U[U[i][h - 1]][h - 1]; 
		
		g = new ArrayList[N + 1];
		for(int i = 0; i < N + 1; i ++)
			g[i] = new ArrayList<Integer>();
		
		L = new ArrayList[N + 1];
		for(int i = 0; i < N + 1; i++)
			L[i] = new ArrayList<Integer>();
		
		for(int i = 1; i < N + 1; i++)
			if(P[i] != -1)g[P[i]].add(i);
		
//		for(int i = 0; i < N + 1; i++)
//			System.out.println(i + ": " + g[i]);
		
		max = new int[N + 1];
		min = new int[N + 1];
		
		dfs(0, 0);
		
//		System.out.println("-----");
//		for(int l = 0; l <= 4; l++)
//			System.out.println(l + ": " + L[l]);
//		
//		System.out.println("-----");
//		
//		for(int i = 0; i < N + 1; i++)
//			System.out.println(i + ": " + H[i]);
		
		for(int i = 0; i < N + 1; i++)
			Collections.sort(L[i]);
		
		int Q = r.nextInt();
		while(Q-- > 0){
			int v = r.nextInt();
			int p = r.nextInt();
			
			int parent = v;
			
			while(p > 0){
				int h = log[p & -p];
				
				parent = U[parent][h];
				if(parent == -1)break;
				
				p -= p & -p;
			}
			
			if(parent == -1 || parent == 0){
				out.println(0);
				continue;
			}
			
//			out.println(v + ", " + parent);
			
			
			int lo = 0, hi = L[H[v]].size() - 1; 
			while(lo < hi){
				int md = (lo + hi) / 2;
				
				if(min[parent] <= L[H[v]].get(md))hi = md;
				else lo = md + 1;
			}
			
			if(min[parent] > L[H[v]].get(lo)){
				out.println(0);
				continue;
			}
			
			int Left = lo;
			
			lo = 0;
			hi = L[H[v]].size() - 1;
			
			while(lo < hi){
				int md = (lo + hi + 1) / 2;
				
				if(L[H[v]].get(md) <= max[parent])lo = md;
				else hi = md - 1;
			}
			
			if(L[H[v]].get(lo) > max[parent]){
				out.println(0);
				continue;
			}
			
			int Right = lo;
			
//			out.println(min[parent] + " -> " + max[parent]);
//			out.println(L[H[v]].get(Left) + ", " + L[H[v]].get(Right));
			out.println(Right - Left + 1 - 1);
			
		}
		
		out.close();
	}
	
	private static void dfs(int i, int l) {
		
		H[i] = l;
		
		M[i] = id++; 
		L[l].add(M[i]);
		min[i] = id;
		for(int j : g[i])
			dfs(j, l + 1);
		max[i] = id - 1;
	}

	static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
				return null;
			}
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}
		
		public double nextDouble(){
			return Double.parseDouble(next());
		}
	}
}